Comprehensive Study Notes on Calculus: Limits, Continuity, Differentiability, Mean Value Theorems, Asymptotes, and Curve Tracing

Limits and Neighbourhoods

• Neighbourhood of a Point:

  • In a space with a distance function $d(\text{--}, \text{--})$, the neighbourhood of a point $a$ with radius ̢\delta (denoted by $N_a(\delta)$) is defined as:     $N_a(\delta) = \{x : d(x, a) < \delta\}$

  • On the real line $\mathbb{R}$, the distance function is $d(x, a) = |x - a|$. Thus, for $a \in \mathbb{R}$:     $N_a(\delta) = \{x : |x - a| < \delta\}$

  • Example 1.1: Finding the neighbourhood of $a = 2$ with $\delta = 1$:     $N_2(1) = \{x : |x - 2| < 1\} = (1, 3)$     This represents the set of all points lying in the open interval $(1, 3)$.     

    

  • Example 1.2: Finding the neighbourhood of $a = -1$ with $\delta = 2$:     $N_{-1}(2) = \{x : |x - (-1)| < 2\} = (-3, 1)$

• Deleted Neighbourhood of a Point:

  • The deleted neighbourhood of a point $a \in \mathbb{R}$ (denoted by $N_a^{*}(\delta)$) excludes the point $a$ itself:     $N_a^{*}(\delta) = \{x : 0 < d(x, a) < \delta\} = \{x : 0 < |x - a| < \delta\}$

  • Example 1.3: Finding the deleted neighbourhood of $a = 2$ with $\delta = 1$:     $N_2^{*}(1) = \{x : 0 < |x - 2| < 1\} = (1, 3) \setminus \{2\} = (1, 2) \cup (2, 3)$     

    

  • Example 1.4: Finding the deleted neighbourhood of $a = \frac{1}{2}$ with $\delta = \frac{3}{4}$:     $N_{1/2}^{*}(3/4) = \left\{x : 0 < \left|x - \frac{1}{2}\right| < \frac{3}{4}\right\} = \left(\frac{1}{2} - \frac{3}{4}, \frac{1}{2} + \frac{3}{4}\right) \setminus \left\{\frac{1}{2}\right\} = \left(-\frac{1}{4}, \frac{1}{2}\right) \cup \left(\frac{1}{2}, \frac{5}{4}\right)$

Formal Definition of a Limit and Epsilon-Delta Proofs

• Motivation:

  • For functions like $f(x) = \frac{x^2 - 4}{x - 2}$, evaluating $f(2)$ directly yields an undefined expression. Understanding the behavior of $f(x)$ when $x$ is extremely close to $2$ requires the concept of a limit.

• Formal Definition of Limit:

  • Let $f : I \subseteq \mathbb{R} \rightarrow \mathbb{R}$ be a real-valued function. A real number $l$ is said to be the limit of $f(x)$ at $x = a$ if for any $\epsilon > 0$, there exists a $\delta > 0$ (depending on $\epsilon$ and $a$) such that for all $x$:     $|f(x) - l| < \epsilon \quad \text{whenever} \quad 0 < |x - a| < \delta$

  • Symbolically written as:     $\lim_{x \rightarrow a} f(x) = l$

  • Alternative representations:     $0 < |x - a| < \delta \implies |f(x) - l| < \epsilon$     $x \in N_a^{*}(\delta) \implies f(x) \in N_l(\epsilon)$

  • Note: $f(a)$ may not exist, or if it exists, $f(a)$ might not belong to $N_l(\epsilon)$.   

    

• Detailed Epsilon-Delta Proof Examples:

  • Example 1.6: Prove $\lim_{x \rightarrow 2} \frac{x^2 - 4}{x - 2} = 4$

  • Estimate: $|f(x) - 4| = \left|\frac{x^2 - 4}{x - 2} - 4\right| = |(x + 2) - 4| = |x - 2|$

  • Choice: Choosing $\delta = \epsilon$ gives $|f(x) - 4| = |x - 2| < \epsilon$ whenever $0 < |x - 2| < \delta$.

  • Example 1.7: Prove $\lim_{x \rightarrow 0} (2x + 5) = 5$

  • Estimate: $|f(x) - 5| = |(2x + 5) - 5| = 2|x|$

  • Choice: Choosing $\delta = \frac{\epsilon}{2}$ gives $|f(x) - 5| = 2|x| < 2 \left(\frac{\epsilon}{2}\right) = \epsilon$ whenever $0 < |x - 0| < \delta$.

  • Example 1.8: Prove $\lim_{x \rightarrow 2} \frac{2(x^2 + x - 6)}{x - 2} = 10$

  • Estimate: $|f(x) - 10| = \left|\frac{2(x - 2)(x + 3)}{x - 2} - 10\right| = |2x - 4| = 2|x - 2|$

  • Choice: Choosing $\delta = \frac{\epsilon}{2}$ gives $|f(x) - 10| = 2|x - 2| < \epsilon$ whenever $0 < |x - 2| < \delta$.

  • Example 1.9: Prove $\lim_{x \rightarrow 2} x^2 = 4$

  • Analysis: $|f(x) - 4| = |x^2 - 4| = |x + 2||x - 2|$. Assuming $\delta < 1$, $-1 < x - 2 < 1 \implies 3 < x + 2 < 5 \implies |x + 2| < 5$.

  • Thus, $|f(x) - 4| < 5|x - 2|$. To ensure $5|x - 2| < \epsilon$, take $|x - 2| < \frac{\epsilon}{5}$.

  • Formal Choice: ${\delta = \min\left(1, \frac{\epsilon}{5}\right)}$. Then $|f(x) - 4| < 5|x - 2| < 5\left(\frac{\epsilon}{5}\right) = \epsilon$ whenever |x - 2| < \delta$.\n * General bound method: |x| = |x - 2 + 2| \le |x - 2| + 2 < \delta + 2 < 3$(assuming$\delta < 1$). Then$|f(x) - 4| < (|x| + 2)|x - 2| < 5|x - 2|.\n * **Example 1.10**: For \lim_{x \rightarrow 5} \sqrt{x - 1} = 2$, find$\delta > 0$for$\epsilon = 1\n * Step (i): Solve |\sqrt{x - 1} - 2| < 1 \iff -1 < \sqrt{x - 1} - 2 < 1 \iff 1 < \sqrt{x - 1} < 3 \iff 1 < x - 1 < 9 \iff 2 < x < 10.\n * Step (ii): Centered interval (5 - \delta, 5 + \delta)$inside$(2, 10)$. Distance from$5$to nearest endpoint is$3$. Thus, choosing$\delta = 3$guarantees$0 < |x - 5| < 3 \implies |\sqrt{x - 1} - 2| < 1$.

  • Example 1.11: Prove $\lim_{x \rightarrow 1} \sqrt{1 + x} = \sqrt{2}$

  • Estimate: $|f(x) - \sqrt{2}| = |\sqrt{1 + x} - \sqrt{2}| = \left|\frac{1 + x - 2}{\sqrt{1 + x} + \sqrt{2}}\right| = \frac{|x - 1|}{\sqrt{1 + x} + \sqrt{2}}$.

  • Since $\sqrt{1 + x} + \sqrt{2} > 1$ on domain $(-1, \infty)$, $|f(x) - \sqrt{2}| < |x - 1|$.

  • Choice: Choosing $\delta = \epsilon$ yields $|f(x) - \sqrt{2}| < \epsilon$ whenever 0 < |x - 1| < \delta$.\n * **Example 1.12**: Prove \lim_{x \rightarrow 1} \sqrt{8 + x^2} = 3\n * Solve |\sqrt{8 + x^2} - 3| < \epsilon \iff 3 - \epsilon < \sqrt{8 + x^2} < 3 + \epsilon \iff 0 \le x^2 < \epsilon^2 + 6\epsilon + 1 \iff -\sqrt{\epsilon^2 + 6\epsilon + 1} < x < \sqrt{\epsilon^2 + 6\epsilon + 1}.\n * Subtract 1$:$-1 - \sqrt{\epsilon^2 + 6\epsilon + 1} < x - 1 < \sqrt{\epsilon^2 + 6\epsilon + 1} - 1$.

  • Choice: ${\delta = \sqrt{\epsilon^2 + 6\epsilon + 1} - 1}$. Then 0 < |x - 1| < \delta \implies |f(x) - 3| < \epsilon$.\n * **Example 1.13**: Prove \lim_{x \rightarrow 0} \frac{x^2 + 2}{x^2 + 1/2} = 4\n * Estimate: |f(x) - 4| = \left|\frac{x^2 + 2}{x^2 + 1/2} - 4\right| = \left|\frac{3x^2}{x^2 + 1/2}\right| < 3$for all$x$.

  • If $\epsilon \ge 3$, $|f(x) - 4| < \epsilon$ holds for all $x$.

  • If $\epsilon < 3$, \frac{3x^2}{x^2 + 1/2} < \epsilon \iff 3x^2 < \epsilon\left(x^2 + \frac{1}{2}\right) \iff x^2 < \frac{\epsilon}{2(3 - \epsilon)} \iff -\sqrt{\frac{\epsilon}{2(3 - \epsilon)}} < x < \sqrt{\frac{\epsilon}{2(3 - \epsilon)}}$.\n * General Choice: {\delta = \sqrt{\frac{\epsilon}{2|3 - \epsilon|}}}.\n * **Example 1.14**: Prove \lim_{x \rightarrow 2} \frac{1}{x} = \frac{1}{2}\n * Estimate: |f(x) - 1/2| = \left|\frac{1}{x} - \frac{1}{2}\right| = \frac{|x - 2|}{2|x|}.\n * Assuming \delta = 1$,$0 < |x - 2| < 1 \implies 1 < x < 3 \implies \frac{1}{|x|} < 1 \implies \frac{|x - 2|}{2|x|} < \frac{|x - 2|}{2}.\n * Choice: {\delta = \min(1, 2\epsilon)}$. Then$|f(x) - 1/2| < \frac{|x - 2|}{2} < \frac{2\epsilon}{2} = \epsilon$.

  • Example 1.15: Prove $\lim_{x \rightarrow 0} x \sin\left(\frac{1}{x}\right) = 0$

  • Estimate: $|f(x) - 0| = \left|x \sin\left(\frac{1}{x}\right)\right| \le |x|$ since $\left|\sin\left(\frac{1}{x}\right)\right| \le 1$.

  • Choice: Choosing $\delta = \epsilon$ gives $|f(x) - 0| < \epsilon$ whenever 0 < |x - 0| < \delta$.\n * **Example 1.16**: Prove \lim_{x \rightarrow 0} \sin\left(\frac{1}{x}\right) does not exist\n * Negation of limit definition: Show that for any candidate l$, there exists an$\epsilon > 0$such that for every$\delta > 0$, there exists$x_1$with$0 < |x_1 - 0| < \delta$but$|f(x_1) - l| \ge \epsilon$.

  • Case 1: $|l| \neq 0$. Set $\epsilon = \frac{|l|}{2}$. For any $\delta > 0$, choose integer $n$ large enough such that $x_1 = \frac{1}{n\pi}$ satisfies $0 < |x_1| < \delta$. Then |\sin(n\pi) - l| = |l| > \frac{|l|}{2} = \epsilon$.\n * Case 2: |l| = 0$. Set$\epsilon = \frac{1}{2}$. For any$\delta > 0$, choose integer$n$large enough such that$x_1 = \frac{1}{2n\pi + \pi/2}$satisfies$0 < |x_1| < \delta$. Then$|\sin(2n\pi + \pi/2) - 0| = 1 > \frac{1}{2} = \epsilon$.

  • Therefore, no real number $l$ can be the limit.

One-Sided Limits and Limit Theorems

• Left-Hand Limit (L.H.L.):

  • $f(x)$ tends to $l$ as $x \rightarrow a^{-}$ if for each $\epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ whenever $a - \delta < x < a$.

  • Denoted by $f(a - 0) = \lim_{x \rightarrow a^{-}} f(x)$.

  • Calculation method: Substitute $x = a - h$ ($h > 0$) and evaluate $\lim_{h \rightarrow 0^{+}} f(a - h)$.

• Right-Hand Limit (R.H.L.):

  • $f(x)$ tends to $l$ as $x \rightarrow a^{+}$ if for each $\epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ whenever $a < x < a + \delta$.

  • Denoted by $f(a + 0) = \lim_{x \rightarrow a^{+}} f(x)$.

  • Calculation method: Substitute $x = a + h$ ($h > 0$) and evaluate $\lim_{h \rightarrow 0^{+}} f(a + h)$.

• Existence of Limit at a Point:

  • $\lim_{x \rightarrow a} f(x)$ exists if and only if both L.H.L. and R.H.L. exist and are equal: f(a - 0) = f(a + 0) = l$.\n\n* **Examples**:\n * **Example 1.17**: Prove \lim_{x \rightarrow 0^{+}} \frac{2}{1 + e^{-1/x}} = 2\n * Estimate: \left|\frac{2}{1 + e^{-1/x}} - 2\right| = \left|\frac{-2e^{-1/x}}{1 + e^{-1/x}}\right| = \frac{2}{e^{1/x} + 1}.\n * For \epsilon \ge 1$, any$\delta > 0$works since$\frac{2}{e^{1/x} + 1} < 1$.

  • For $0 < \epsilon < 1$: $\frac{2}{e^{1/x} + 1} < \epsilon \iff e^{1/x} > \frac{2}{\epsilon} - 1 \iff \frac{1}{x} > \ln\left(\frac{2}{\epsilon} - 1\right) \iff 0 < x < \frac{1}{\ln\left(\frac{2}{\epsilon} - 1\right)}$.

  • Choice: ${\delta = \frac{1}{\ln\left(\frac{2}{\epsilon} - 1\right)}}$.

  • Example 1.18: Show $\lim_{x \rightarrow 1} [x]$ does not exist ($[x]$ is greatest integer function)

  • L.H.L.: $\lim_{x \rightarrow 1^{-}} [x] = \lim_{h \rightarrow 0^{+}} [1 - h] = 0$

  • R.H.L.: $\lim_{x \rightarrow 1^{+}} [x] = \lim_{h \rightarrow 0^{+}} [1 + h] = 1$

  • Since L.H.L. $\neq$ R.H.L., the limit does not exist.

  • Example 1.19: Evaluate $\lim_{x \rightarrow 3} \left([x] - \left[\frac{x}{3}\right]\right)$

  • Near $x = 3$:

    • For $2 < x < 3$: $[x] = 2$, $\left[\frac{x}{3}\right] = 0 \implies f(x) = 2 - 0 = 2$

    • For $3 \le x < 4$: $[x] = 3$, $\left[\frac{x}{3}\right] = 1 \implies f(x) = 3 - 1 = 2$

  • L.H.L. = $2$, R.H.L. = $2$. Thus, \lim_{x \rightarrow 3} f(x) = 2$.\n * **Example 1.20**: Show \lim_{x \rightarrow 1} \sin\left(\frac{1}{x - 1}\right) does not exist\n * L.H.L. = \lim_{h \rightarrow 0^{+}} \sin\left(\frac{1}{1 - h - 1}\right) = -\lim_{h \rightarrow 0^{+}} \sin\left(\frac{1}{h}\right).\n * As h \rightarrow 0^{+}$,$\sin\left(\frac{1}{h}\right)$oscillates finitely between$-1$and$1, failing to approach a unique value. L.H.L. and R.H.L. do not exist.\n\n* **Theorem 1.1 (Limit Laws)**:\n * Let f, g : [b, c] \rightarrow \mathbb{R}$with$\lim_{x \rightarrow a} f(x) = l$and$\lim_{x \rightarrow a} g(x) = m:\n 1. \lim_{x \rightarrow a} (f \pm g)(x) = l \pm m\n 2. \lim_{x \rightarrow a} (kf)(x) = kl$for any$k \in \mathbb{R}\n 3. \lim_{x \rightarrow a} (f \cdot g)(x) = l \cdot m\n 4. \lim_{x \rightarrow a} \left(\frac{f}{g}\right)(x) = \frac{l}{m}$provided$g(x) eq 0$in$N_a^{*}(\delta)$and$m eq 0\n 5. \lim_{x \rightarrow a} [f(x)]^n = l^n$($n \in \mathbb{Z}^+)\n 6. \lim_{x \rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{l} = l^{1/n}$($n \in \mathbb{Z}^+)\n\n\n# Infinite Limits\n\n* **Formal Definitions**:\n * \lim_{x \rightarrow a} f(x) = \infty$: For any$G > 0$, there exists$\delta > 0$such that$f(x) > G$whenever$0 < |x - a| < \delta.\n * \lim_{x \rightarrow a^{+}} f(x) = \infty$: For any$G > 0$, there exists$\delta > 0$such that$f(x) > G$whenever$a < x < a + \delta.\n * \lim_{x \rightarrow a^{-}} f(x) = \infty$: For any$G > 0$, there exists$\delta > 0$such that$f(x) > G$whenever$a - \delta < x < a.\n * \lim_{x \rightarrow a} f(x) = -\infty$: For any$G > 0$, there exists$\delta > 0$such that$f(x) < -G$whenever$0 < |x - a| < \delta.\n\n* **Examples**:\n * **Example 1.21**: Prove \lim_{x \rightarrow 0} \frac{1}{x} does not exist\n * Right limit: For G > 0$, set$\delta = \frac{1}{G}$.$0 < x < \delta \implies \frac{1}{x} > G \implies \lim_{x \rightarrow 0^{+}} \frac{1}{x} = \infty$.

  • Left limit: For $G > 0$, set $\delta = \frac{1}{G}$. -\delta < x < 0 \implies \frac{1}{x} < -G \implies \lim_{x \rightarrow 0^{-}} \frac{1}{x} = -\infty$.\n * Since L.H.L. eq R.H.L., the limit does not exist.\n * **Example 1.22**: Prove \lim_{x \rightarrow 0} \frac{1}{x^2} = \infty\n * For G > 0$,$f(x) > G \iff \frac{1}{x^2} > G \iff x^2 < \frac{1}{G} \iff -\frac{1}{\sqrt{G}} < x < \frac{1}{\sqrt{G}}.\n * Choose {\delta = \frac{1}{\sqrt{G}}}$. Then$0 < |x - 0| < \delta \implies f(x) > G$.

  • Example 1.23: Prove $\lim_{x \rightarrow 0} \log|x| = -\infty$

  • For $G > 0$, $f(x) < -G \iff \log|x| < -G \iff |x| < \exp(-G)$.

  • Choose ${\delta = \exp(-G)}$. Then 0 < |x - 0| < \delta \implies f(x) < -G$.\n * **Example 1.24**: Prove \lim_{x \rightarrow 1} f(x) = -\infty where:\n    f(x) = \begin{cases} \frac{1}{x - 1}, & x < 1 \ \log(x - 1), & x > 1 \end{cases}\n * For x < 1$:$\frac{1}{x - 1} < -G \iff x - 1 > -\frac{1}{G} \iff 0 < 1 - x < \frac{1}{G} \iff 0 < |x - 1| < \frac{1}{G}.\n * For x > 1$:$\log(x - 1) < -G \iff x - 1 < e^{-G} \iff 0 < |x - 1| < e^{-G}.\n * Choose {\delta = \min\left(\frac{1}{G}, e^{-G}\right)}$. Then$0 < |x - 1| < \delta \implies f(x) < -G$.

• Practice Problems:

  1. Find $\delta$ such that $0 < |x - 2| < \delta \implies \left|\frac{2x^3 + x^2 - 8x - 4}{x - 2} - 20\right| < \frac{1}{10}$.

  2. Show $\lim_{x \rightarrow 3} \frac{\sqrt{x}}{1 + 2x} = \frac{\sqrt{3}}{7}$ using $\epsilon - \delta$.

  3. Show $\lim_{x \rightarrow 1} \frac{2x^4 - 6x^3 + x^2 + 3}{x - 1} = -8$ using $\epsilon - \delta$.

  4. For $f(x) = x + 1$ ($x > 0$), $0$ ($x = 0$), $x - 1$ ($x < 0$), show $\lim_{x \rightarrow 0} f(x)$ does not exist but one-sided limits exist.

  5. Show $\lim_{x \rightarrow a} \frac{1}{(x - a)^n} = \infty$ for even natural $n$, and does not exist for odd natural $n$

  6. Show $\lim_{x \rightarrow 1} 2^{\frac{1}{x - 1}}$ does not exist.

Continuity and Discontinuity

• Definition of Continuity:

  • A function $f(x)$ is continuous at $x = a$ if for any $\epsilon > 0$, there exists $\delta > 0$ such that:     $|f(x) - f(a)| < \epsilon \quad \text{whenever} \quad |x - a| < \delta$

  • Equivalently: $\lim_{x \rightarrow a} f(x) = f(a)$.

  • Three formal conditions:

  1. $f(a)$ is defined.

  2. $\lim_{x \rightarrow a} f(x)$ exists.

  3. $\lim_{x \rightarrow a} f(x) = f(a)$.

• One-Sided Continuity & Interval Continuity:

  • Continuous from left at $a$: $\lim_{x \rightarrow a^{-}} f(x) = f(a)$.

  • Continuous from right at $a$: $\lim_{x \rightarrow a^{+}} f(x) = f(a)$.

  • Continuity on endpoint of $[a, b]$: Right-continuous at $a$ ($\lim_{x \rightarrow a^{+}} f(x) = f(a)$), left-continuous at $b$ ($\lim_{x \rightarrow b^{-}} f(x) = f(b)$).

  • Continuous on interval $[a, b]$: Continuous at every point in $[a, b]$.

• Examples of Continuity Proofs:

  • Example 2.1: Prove $f(x) = \sin(x)$ is continuous at $x = 0$

  • $|f(x) - f(0)| = |\sin(x) - 0| = |\sin(x)| \le |x| = |x - 0|$.

  • Choose $\delta = \epsilon$. Then |x - 0| < \delta \implies |f(x) - f(0)| < \epsilon$.\n * **Example 2.2**: Prove f(x) = e^x$is continuous at$x = 0\n * Solve |e^x - 1| < \epsilon \iff 1 - \epsilon < e^x < 1 + \epsilon \implies -\log(1 + \epsilon) < x < \log(1 + \epsilon).\n * Choose {\delta = \log(1 + \epsilon)}.\n * **Example 2.3**: Prove f(x) = \log(x)$is continuous at$x = 2\n * Estimate: |\log(x) - \log(2)| = \left|\log\left(\frac{x}{2}\right)\right| = \left|\log\left(\frac{x - 2}{2} + 1\right)\right| \le \left|\frac{x - 2}{2}\right|$(assuming$\delta < 1).\n * Choose {\delta = \min(1, 2\epsilon)}$. Then$|\log(x) - \log(2)| \le \frac{|x - 2|}{2} < \epsilon$.

  • Example 2.4: Prove continuity at $x = 1/2$ for $f(x) = x$ (rational) and $1 - x$ (irrational)

  • $|f(x) - f(1/2)| = \left|x - \frac{1}{2}\right|$ if rational, and $\left|(1 - x) - \frac{1}{2}\right| = \left|\frac{1}{2} - x\right| = \left|x - \frac{1}{2}\right|$ if irrational.

  • Thus for all $x$, $|f(x) - f(1/2)| = |x - 1/2|$. Choose \delta = \epsilon$.\n * **Example 2.5**: Prove f(x) = x^2$is continuous on$\mathbb{R}\n * Fix a \in \mathbb{R}$.$|x^2 - a^2| = |x + a||x - a| \le (|x| + |a|)|x - a|.\n * Assuming \delta = 1$,$|x| < 1 + |a| \implies |x^2 - a^2| < (1 + 2|a|)|x - a|$.

  • Choose ${\delta = \min\left(1, \frac{\epsilon}{1 + 2|a|}\right)}$.

• Types of Discontinuities:

  • Removable Discontinuity: $\lim_{x \rightarrow a} f(x)$ exists, but $\lim_{x \rightarrow a} f(x) \neq f(a)$.     

    

  • Jump Discontinuity (First Kind): Both $\lim_{x \rightarrow a^{-}} f(x)$ and $\lim_{x \rightarrow a^{+}} f(x)$ exist, but are not equal.     

    

  • Oscillatory Discontinuity (Second Kind): Neither one-sided limit exists.     

    

  • Infinite Discontinuity: $\lim_{x \rightarrow a} f(x) = \infty$ or $-\infty$.     

    

        

    

• Discontinuity Examples:

  • Example 2.6: $f(x) = \frac{\sin(2x)}{x}$ ($x \neq 0$), $5$ ($x = 0$). $\lim_{x \rightarrow 0} f(x) = 2 \neq 5 \implies$ Removable discontinuity.

  • Example 2.7: $f(x) = \frac{x^2 - 4}{x - 2}$ ($x \neq 2$), $10$ ($x = 2$). $\lim_{x \rightarrow 2} f(x) = 4 \neq 10 \implies$ Removable discontinuity.

  • Example 2.8: $f(x) = [x]$ at $x = 1$. L.H.L. = $0$, R.H.L. = $1 \implies$ Jump discontinuity.

  • Example 2.9: $f(x) = \frac{x - |x|}{x}$ ($x \neq 0$), $2$ ($x = 0$). L.H.L. = $2$, R.H.L. = $0$, $f(0) = 2 \implies$ Right jump discontinuity at x = 0$.\n * **Example 2.10**: f(x) = x - [x]$at$x = 3$. L.H.L. =$1$, R.H.L. =$0$,$f(3) = 0 \implies Left jump discontinuity.\n * **Example 2.11**: f(x) = \sin\left(\frac{1}{x}\right)$at$x = 0 \implies Oscillatory discontinuity.\n * **Example 2.12**: f(x) = \frac{1}{x^2}$at$x = 0 \implies Infinite discontinuity.\n\n* **Theorem 2.1 (Algebra of Continuous Functions)**:\n * If f, g$are continuous at$x = a$, then$f \pm g$,$kf$,$f \cdot g$,$f/g$($g(a) eq 0$),$f^n$, and$\sqrt[n]{f}$are continuous at$x = a\n\n\n# Composition of Continuous Functions\n\n* **Theorem**:\n * If f$is continuous at$a$and$g$is continuous at$f(a)$, then the composite function$(g \circ f)(x) = g(f(x))$is continuous at$a\n\n* **Examples**:\n * **Example 2.13(a)**: h_1(x) = \sqrt{x^2 - 2x - 5}\n * Polynomial f(x) = x^2 - 2x - 5 is everywhere continuous.\n * Square root g(x) = \sqrt{x}$is continuous on$[0, \infty).\n * Composite (g \circ f)(x) = \sqrt{x^2 - 2x - 5}$is continuous on its domain where$x^2 - 2x - 5 \ge 0$.

  • Example 2.13(b): $h_2(x) = \sin(x^2)$

  • Polynomial $f(x) = x^2$ and trigonometric function $g(x) = \sin(x)$ are everywhere continuous.

  • Composite $(g \circ f)(x) = \sin(x^2)$ is everywhere continuous.

• Exercise Problems:

  1. Prove continuity of: (a) $|x|$ over $\mathbb{R}$, (b) $\sin(x), \cos(x)$ over $\mathbb{R}$, (c) $x^n$ on $\mathbb{R}$ ($n \in \mathbb{Q}$), (d) $e^x$ over $\mathbb{R}$, (e) $\tan(x)$ over $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, (f) $\log(x)$ over $(0, \infty)$.

  2. Show continuity everywhere on domain for: (a) $h_3(x) = \frac{x^{2/3}}{1 + x^4}$, (b) $h_4(x) = \left|\frac{x - 2}{x^2 - 2}\right|$, (c) $h_5(x) = \left|\frac{x \sin(x)}{x^2 + 2}\right|$.

Differentiability and Derivative Rules

• Definition of Differentiability:

  • A function $f(x)$ is differentiable at $x = a$ if the limit exists finitely:     $f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$

• Differentiability Proof Examples:

  • Example 3.1: Constant function $f(x) = k$

  • $\frac{f(x) - f(a)}{x - a} = \frac{k - k}{x - a} = 0$. $\left|0 - 0\right| < \epsilon$ for any $\delta > 0$. Derivative f'(a) = 0$.\n * **Example 3.2**: f(x) = x^2\n * \frac{x^2 - a^2}{x - a} = x + a$.$\left|(x + a) - 2a\right| = |x - a| < \epsilon$by choosing$\delta = \epsilon$. Derivative$f'(a) = 2a$.

  • Example 3.3: $f(x) = \sin(x)$

  • $\frac{\sin(x) - \sin(a)}{x - a} - \cos(a) = \frac{\sin\left(\frac{x-a}{2}\right) \cos\left(\frac{x+a}{2}\right)}{\frac{x-a}{2}} - \cos(a)$.

  • Bound estimate gives $\le \left|\frac{x - a}{2}\right|$. Choosing $\delta = 2\epsilon$ proves f'(a) = \cos(a)$.\n * **Example 3.4**: f(x) = \sqrt{x}$at$x = 3\n * \frac{\sqrt{x} - \sqrt{3}}{x - 3} = \frac{1}{\sqrt{x} + \sqrt{3}}.\n * \left|\frac{1}{\sqrt{x} + \sqrt{3}} - \frac{1}{2\sqrt{3}}\right| = \frac{|x - 3|}{2\sqrt{3}(\sqrt{x} + \sqrt{3})^2} < \frac{|x - 3|}{6\sqrt{3}}.\n * Choosing {\delta = 6\sqrt{3}\epsilon}$yields derivative$f'(3) = \frac{1}{2\sqrt{3}}.\n\n* **Theorem 3.1 (Algebra of Derivatives)**:\n * (f \pm g)'(a) = f'(a) \pm g'(a)\n * (kf)'(a) = k f'(a)\n * (f \cdot g)'(a) = g(a)f'(a) + f(a)g'(a)\n * \left(\frac{f}{g}\right)'(a) = \frac{g(a)f'(a) - f(a)g'(a)}{[g(a)]^2}$(where$g(a) eq 0)\n\n\n# Successive Differentiation and Leibnitz's Theorem\n\n* **Higher-Order Derivatives**:\n * f^{(n)}(x)$or$y_n$or$\frac{d^n y}{dx^n}$denotes the$n$-th derivative of$y = f(x).\n\n* **Worked Examples**:\n * **Example 3.5**: If y = e^{ax} \sin(bx)$, prove$y_2 - 2ay_1 + (a^2 + b^2)y = 0\n * y_1 = a e^{ax} \sin(bx) + b e^{ax} \cos(bx) = a y + b e^{ax} \cos(bx)\n * y_1 - ay = b e^{ax} \cos(bx)\n * Differentiating again: y_2 - ay_1 = ab e^{ax} \cos(bx) - b^2 e^{ax} \sin(bx) = a(y_1 - ay) - b^2 y\n * Rearranging gives y_2 - 2ay_1 + (a^2 + b^2)y = 0$.

  • Example 3.6: If $y = \frac{1}{x^2 - a^2}$, find $y_n$ and $y_n(0)$

  • Partial fractions: $y = \frac{1}{2a}\left(\frac{1}{x - a} - \frac{1}{x + a}\right)$

  • $y_n = \frac{(-1)^n n!}{2a}\left[\frac{1}{(x - a)^{n+1}} - \frac{1}{(x + a)^{n+1}}\right]$

  • Evaluating at $x = 0$:       $y_n(0) = \begin{cases} \frac{(-1)^{n+1} n!}{a^{n+2}}, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases}$

  • Example 3.7: If $y = x \log\left(\frac{x - 1}{x + 1}\right)$, prove $y_n = (-1)^{n-2}(n-2)! \left[\frac{x - n}{(x - 1)^n} - \frac{x + n}{(x + 1)^n}\right]$ for $n \ge 2$, and find $y_n(0)$

  • $y_1 = \log(x - 1) - \log(x + 1) + \frac{1}{x - 1} + \frac{1}{x + 1}$

  • Differentiating $(n-1)$ times yields the required $y_n$ expression.

  • Evaluating at $x = 0$:       $y_n(0) = \begin{cases} (-1)^{n-1} 2n (n - 2)!, & \text{if } n \text{ is even} \\ 0, & \text{if } n \text{ is odd} \end{cases}$

• Theorem 3.2 (Leibnitz's Theorem):

  • For product of two functions $(f \cdot g)_n(x) = \sum_{i=0}^{n} \binom{n}{i} f_i(x) g_{n-i}(x)$

• Leibnitz's Theorem Applications:

  • Example 3.8: If $y = (\arcsin(x))^2$, prove $(1 - x^2)y_{n+2} - (2n + 1)x y_{n+1} - n^2 y_n = 0$ for $n \ge 1$, and evaluate $y_n(0)$

  • $y_1 = \frac{2\arcsin(x)}{\sqrt{1 - x^2}} \implies (1 - x^2)y_1^2 = 4y$

  • Differentiating: $(1 - x^2)2y_1 y_2 - 2x y_1^2 = 4y_1 \implies (1 - x^2)y_2 - x y_1 = 2$

  • Applying Leibnitz's theorem $n$ times:       $(1 - x^2)y_{n+2} + n(-2x)y_{n+1} + \frac{n(n-1)}{2}(-2)y_n - x y_{n+1} - n y_n = 0$       $(1 - x^2)y_{n+2} - (2n + 1)x y_{n+1} - n^2 y_n = 0$

  • At $x = 0$: $y_{n+2}(0) = n^2 y_n(0)$.

  • Initial values: $y(0) = 0$, $y_1(0) = 0$, $y_2(0) = 2$

  • Results:

    • If $n$ is odd: $y_n(0) = 0$

    • If $n$ is even ($n > 2$): $y_n(0) = (n - 2)^2 (n - 4)^2 \cdots 6^2 \cdot 4^2 \cdot 2^2 \cdot 2$

  • Example 3.9: If $y^{1/m} + y^{-1/m} = 2x$, prove $(x^2 - 1)y_{n+2} + (2n + 1)x y_{n+1} + (n^2 - m^2)y_n = 0$

  • Quadratic in $y^{1/m}$: $(y^{1/m})^2 - 2x y^{1/m} + 1 = 0 \implies y = \left[x \pm \sqrt{x^2 - 1}\right]^m$

  • Logarithmic differentiation: $\log(y) = m \log\left[x \pm \sqrt{x^2 - 1}\right] \implies y_1\sqrt{x^2 - 1} = \pm m y$

  • Square and differentiate: $(x^2 - 1)y_1^2 = m^2 y^2 \implies (x^2 - 1)y_2 + x y_1 = m^2 y$

  • Apply Leibnitz's theorem $n$ times: (x^2 - 1)y_{n+2} + (2n + 1)x y_{n+1} + (n^2 - m^2)y_n = 0$.\n\n\n# Rolle's Theorem and Mean Value Theorems\n\n* **Theorem 4.1 (Rolle's Theorem)**:\n * Let f : [a, b] \rightarrow \mathbb{R}$be continuous on$[a, b]$and differentiable on$(a, b)$. If$f(a) = f(b)$, then there exists$c \in (a, b)$such that$f'(c) = 0$.   

    

  • Example 4.1: Verify Rolle's theorem for $f(x) = 2x^3 + x^2 - 4x - 2$

  • Roots of $f(x) = 0$: $(x^2 - 2)(2x + 1) = 0 \implies x = -1/2, \pm\sqrt{2}$.

  • Choose interval $[-\sqrt{2}, \sqrt{2}]$. $f(-\sqrt{2}) = f(\sqrt{2}) = 0$.

  • $f'(c) = 6c^2 + 2c - 4 = 0 \implies (3c - 2)(c + 1) = 0 \implies c = 2/3, -1$.

  • Both $c = 2/3$ and $c = -1$ lie within $(-\sqrt{2}, \sqrt{2})$.

  • Example 4.2: Prove that between any real roots of polynomial $p(x) = 0$ (degree $> 1$), there is a real root of $p'(x) + k p(x) = 0$

  • Set $f(x) = e^{kx} p(x)$. Let $\alpha < \beta$ be real roots of $p(x)$.

  • $f(\alpha) = e^{k\alpha} p(\alpha) = 0 = e^{k\beta} p(\beta) = f(\beta)$.

  • By Rolle's theorem on $[\alpha, \beta]$, $f'(c) = 0 \implies e^{kc} p'(c) + k e^{kc} p(c) = 0 \implies p'(c) + k p(c) = 0$ ($e^{kc} \neq 0$).

• Theorem 4.2 (Lagrange's Mean Value Theorem):

  • Let $f : [a, b] \rightarrow \mathbb{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists $c \in (a, b)$ such that:     $\frac{f(b) - f(a)}{b - a} = f'(c)$   

    

  • Example 4.3: Prove $\frac{b - a}{1 + b^2} < \arctan(b) - \arctan(a) < \frac{b - a}{1 + a^2}$ for $0 < a < b < 1$

  • Let $f(x) = \arctan(x)$ on $[a, b]$. LMV'T gives $\frac{\arctan(b) - \arctan(a)}{b - a} = \frac{1}{1 + c^2}$ for a < c < b$.\n * Since a < c < b \implies 1 + a^2 < 1 + c^2 < 1 + b^2 \implies \frac{1}{1 + b^2} < \frac{1}{1 + c^2} < \frac{1}{1 + a^2}.\n * Multiplying by b - a yields the inequality.\n\n* **Theorem 4.3 (Cauchy's Mean Value Theorem)**:\n * Let f, g : [a, b] \rightarrow \mathbb{R}$be continuous on$[a, b]$and differentiable on$(a, b)$. If$g'(x) eq 0$, then there exists$c \in (a, b) such that:\n    \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}\n  \n * **Example 4.4**: Prove that c$is the arithmetic mean of$a$and$b$for$f(x) = e^x, g(x) = e^{-x}$on$[a, b]\n * \frac{e^b - e^a}{e^{-b} - e^{-a}} = \frac{e^c}{-e^{-c}} \implies \frac{e^b(1 - e^{a-b})}{e^{-a}(e^{a-b} - 1)} = -e^{2c} \implies e^{a+b} = e^{2c} \implies c = \frac{a + b}{2}.\n\n\n# Taylor's Theorem and Infinite Series Expansions\n\n* **Theorem 4.4 & 4.5 (Taylor's Theorem with Lagrange's Remainder)**:\n * If f(x)$has continuous derivatives up to$n$-th order in interval$I$containing$a:\n    f(x) = f(a) + (x - a)f'(a) + \frac{(x - a)^2}{2!} f''(a) + \cdots + \frac{(x - a)^{n-1}}{(n - 1)!} f^{(n-1)}(a) + R_n(x)\n    where R_n(x) = \frac{(x - a)^n}{n!} f^{(n)}(c)$for$a < c < x$.

  • Alternative form ($h = x - a$, $c = a + \theta h$, $0 < \theta < 1$):     $f(a + h) = f(a) + h f'(a) + \frac{h^2}{2!} f''(a) + \cdots + \frac{h^{n-1}}{(n - 1)!} f^{(n-1)}(a) + \frac{h^n}{n!} f^{(n)}(a + \theta h)$

• Applications of Taylor's Remainder:

  • Example 4.5: Prove $\frac{x}{1 + x} < \log(1 + x) < x$ for $x > 0$

  • Apply Taylor's theorem to $f(x) = \log(1 + x)$ on $[0, x_0]$: $\log(1 + x_0) = 0 + x_0 \left(\frac{1}{1 + \theta x_0}\right)$.

  • Since 0 < \theta < 1 \implies 1 < 1 + \theta x_0 < 1 + x_0 \implies \frac{x_0}{1 + x_0} < \frac{x_0}{1 + \theta x_0} < x_0$.\n * **Example 4.6**: Prove 0 < \frac{1}{x} \log\left(\frac{e^x - 1}{x}\right) < 1$for$x > 0\n * Apply Taylor's theorem to f(x) = e^x$on$[0, x_0]$:$e^{x_0} = 1 + x_0 e^{\theta x_0} \implies \frac{e^{x_0} - 1}{x_0} = e^{\theta x_0} \implies \frac{1}{x_0} \log\left(\frac{e^{x_0} - 1}{x_0}\right) = \theta$.

  • Since $0 < \theta < 1$, the inequality holds.

  • Example 4.7: Prove $x - \frac{x^3}{6} \le \sin(x) \le x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x > 0$

  • Taylor expansion of $\sin(x)$ with remainder after 3 terms: $\sin(x) = x - \frac{x^3}{3!} \cos(\theta x)$. Since \cos(\theta x) \le 1 \implies \sin(x) \ge x - \frac{x^3}{6}$.\n * With remainder after 5 terms: \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} \cos(\theta_1 x) \le x - \frac{x^3}{6} + \frac{x^5}{120}$.

• Theorem 4.6 & 4.7 (Taylor's and Maclaurin's Infinite Series):

  • As $n \rightarrow \infty$, if $R_n(x) \rightarrow 0$:

  • Taylor Infinite Series at $x = a$:     $f(x) = \sum_{n=0}^{\infty} \frac{(x - a)^n}{n!} f^{(n)}(a)$

  • Maclaurin Infinite Series (at $x = 0$):     $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} f^{(n)}(0)$

• Worked Example & Standard Maclaurin Expansions:

  • Example 4.8: Expand $e^{\sin(x)}$ at $\frac{\pi}{2}$ up to fifth derivative

  • $f(\pi/2) = e$, $f'(\pi/2) = 0$, $f''(\pi/2) = -e$, $f'''(\pi/2) = 0$, $f^{(4)}(\pi/2) = 4e$, f^{(5)}(\pi/2) = 0$.\n * e^{\sin(x)} = e - \frac{e}{2!}\left(x - \frac{\pi}{2}\right)^2 + \frac{4e}{4!}\left(x - \frac{\pi}{2}\right)^4 + \cdots\n * **Standard Maclaurin Expansions**:\n 1. \frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots$($|x| < 1)\n 2. \frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots$($|x| < 1)\n 3. \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\n 4. \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\n 5. \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\n 6. \tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$($|x| < \frac{\pi}{2})\n 7. \sec(x) = 1 + \frac{x^2}{2!} + \frac{5x^4}{4!} + \cdots\n 8. e^x \cos(x) = 1 + x - \frac{2x^3}{3!} - \frac{2^2 x^4}{4!} + \cdots\n 9. \arcsin(x) = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots$($|x| < 1)\n 10. \arccos(x) = \frac{\pi}{2} - x - \frac{x^3}{6} - \frac{3x^5}{40} - \cdots$($|x| < 1)\n 11. \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$($|x| < 1)\n 12. e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\n 13. \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\n 14. \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\n 15. \tanh(x) = x - \frac{x^3}{3} + \frac{2x^5}{15} - \frac{17x^7}{315} + \cdots\n\n\n# Asymptotes of Plane Curves\n\n* **Limits Involving Infinity**:\n * \lim_{x \rightarrow \infty} f(x) = L$if for every$\epsilon > 0$, there exists$M$such that$x > M \implies |f(x) - L| < \epsilon$.

  • $\lim_{x \rightarrow -\infty} f(x) = L$ if for every $\epsilon > 0$, there exists $N$ such that x < N \implies |f(x) - L| < \epsilon$.\n * **Example 5.1**: \lim_{x \rightarrow \infty} \frac{1}{x} = 0$($M = \frac{1}{\epsilon}$),$\lim_{x \rightarrow -\infty} \frac{1}{x} = 0$($N = -\frac{1}{\epsilon}).\n\n* **Asymptotes for Explicit Curves y = f(x)**:\n * **Vertical Asymptote**: Line x = a$if$\lim_{x \rightarrow a^{+}} f(x) = \pm\infty$or$\lim_{x \rightarrow a^{-}} f(x) = \pm\infty.\n * For f(x) = \frac{p(x)}{q(x)}$, vertical asymptotes occur at roots of$q(x) = 0.\n * Odd order root: Curve approaches opposite ends on either side.\n * Even order root: Curve approaches same end on either side.\n * **Horizontal Asymptote**: Line y = b$if$\lim_{x \rightarrow \infty} f(x) = b$or$\lim_{x \rightarrow -\infty} f(x) = b$.

  • Oblique / Inclined Asymptote: Line $y = mx + c$ ($m \neq 0$) if \lim_{x \rightarrow \pm\infty} [f(x) - (mx + c)] = 0$.\n * Slopes and intercepts: m = \lim_{x \rightarrow \pm\infty} \frac{f(x)}{x}$,$c = \lim_{x \rightarrow \pm\infty} [f(x) - mx]$.

• Asymptotes for Implicit Curves $f(x, y) = 0$:

  • Curve equation: a_0 x^n + (a_1 x^{n-1}y + b_1 x^{n-1}) + \cdots = 0 \implies x^n \xi_n(y) + x^{n-1} \xi_{n-1}(y) + \cdots = 0$.\n * **Parallel to X-axis**: Equate the coefficient of the highest degree term in x$to zero ($\xi_n(y) = 0).\n * **Parallel to Y-axis**: Equate the coefficient of the highest degree term in y to zero.\n * **Oblique Asymptotes (y = mx + c)**:\n * Substitute x = 1$,$y = m$into degree$r$terms to form$\phi_r(m).\n * Solve \phi_n(m) = 0$to find values of$m$.

  • For distinct $m$: $c = -\frac{\phi_{n-1}(m)}{\phi_n'(m)}$.

  • For two equal roots $m$: \frac{c^2}{2!} \phi_n''(m) + c \phi_{n-1}'(m) + \phi_{n-2}(m) = 0$.\n * For three equal roots m$:$\frac{c^3}{3!} \phi_n'''(m) + \frac{c^2}{2!} \phi_{n-1}''(m) + c \phi_{n-2}'(m) + \phi_{n-3}(m) = 0$.   

    

• Worked Examples on Asymptotes:

  • Example 5.3: Find asymptotes of $x^2 y^2 - x^2 y - x y^2 + x + y + 1 = 0$

  • Highest power of $x$ is $x^2$, coefficient $y^2 - y = 0 \implies y = 0, y = 1$ (Horizontal).

  • Highest power of $y$ is $y^2$, coefficient $x^2 - x = 0 \implies x = 0, x = 1$ (Vertical).

  • $\phi_4(m) = m^2 = 0 \implies m = 0, 0$ (No inclined asymptotes).

  • Example 5.4: Find asymptotes of $x^2 + 3xy + 2y^2 + 3x - 2y + 1 = 0$

  • Coefficients of $x^2$ and $y^2$ are constant $1$ and $2$ (No parallel asymptotes).

  • \phi_2(m) = 2m^2 + 3m + 1 = 0 \implies m = -1, -1/2$.\n * \phi_1(m) = 3 - 2m$,$\phi_2'(m) = 4m + 3.\n * For m = -1$:$c = -\frac{3 - 2(-1)}{4(-1) + 3} = 5 \implies y = -x + 5$.

  • For $m = -1/2$: c = -\frac{3 - 2(-1/2)}{4(-1/2) + 3} = -4 \implies y = -\frac{1}{2}x - 4$.\n * **Example 5.5(i)**: y^3 - 2xy^2 - x^2y + 2x^3 + 3y^2 - 7xy + 2x^2 + 2y + 2x + 1 = 0\n * \phi_3(m) = m^3 - 2m^2 - m + 2 = (m - 2)(m^2 - 1) = 0 \implies m = 2, 1, -1$.

  • $\phi_2(m) = 3m^2 - 7m + 2$, $\phi_3'(m) = 3m^2 - 4m - 1$.

  • $m = 2 \implies c = 0 \implies y = 2x$

  • $m = 1 \implies c = -1 \implies y = x - 1$

  • $m = -1 \implies c = -2 \implies y = -x - 2$

  • Example 5.5(ii): $x^3 + 3x^2y - 4y^3 - x + y + 3 = 0$

  • \phi_3(m) = 1 + 3m - 4m^3 = (1 - m)(2m + 1)^2 = 0 \implies m = 1, -1/2, -1/2$.\n * Distinct root m = 1 \implies c = 0 \implies y = x$.

  • Double root $m = -1/2$: \frac{c^2}{2}(-24m) + c(0) + (-1 + m) = 0 \implies -6c^2 + 3/2 = 0 \implies c = \pm 1/2$.\n * Asymptotes: y = x$,$y = -\frac{1}{2}x + \frac{1}{2}$,$y = -\frac{1}{2}x - \frac{1}{2}.\n * **Example 5.5(iii)**: (x + y)^2 (x + y + 2) = x + 9y - 2\n * \phi_3(m) = (1 + m)^3 = 0 \implies m = -1, -1, -1 (Triple root).\n * Formula: \frac{c^3}{6}(6) + \frac{c^2}{2}(4) + c(-9) + 2 = 0 \implies c^3 + 2c^2 - 9c + 2 = 0 \implies c = 2, -2 \pm \sqrt{5}$.

  • Asymptotes: $y = -x + 2$, $y = -x - 2 + \sqrt{5}$, y = -x - 2 - \sqrt{5}$.\n\n\n# Curve Tracing in Cartesian Coordinates\n\n* **10-Step Procedure for Cartesian Curve Tracing**:\n 1. **Symmetry**:\n * X-axis: Even powers of y$($f(x, -y) = f(x, y)).\n * Y-axis: Even powers of x$($f(-x, y) = f(x, y)).\n * Origin / Opposite Quadrants: f(-x, -y) = f(x, y).\n * Line y = x$: Interchange$x$and$y$($f(y, x) = f(x, y)).\n * Line y = -x$:$f(-y, -x) = f(x, y).\n 2. **Nature of Origin & Tangents**:\n * Pass through origin if constant term is absent.\n * Tangents at origin obtained by equating lowest degree terms to zero.\n * **Node**: Two real and distinct tangents.\n       \n * **Cusp**: Two real and coincident tangents.\n       \n * **Isolated Point**: Imaginary tangents.\n 3. **Points of Intersection / Intercepts**:\n * X-intercept (y = 0$), Y-intercept ($x = 0).\n 4. **Asymptotes** (Parallel and Oblique).\n 5. **Tangents at Other Points**: Evaluate \frac{dy}{dx}$. Horizontal tangent ($\frac{dy}{dx} = 0$), Vertical tangent ($\frac{dy}{dx} = \infty).\n 6. **Region of Existence / Absence**: Values where y becomes imaginary.\n 7. **Intervals of Increase/Decrease**: Sign of f'(x).\n 8. **Local Extrema**: Critical points and sign of f''(x).\n 9. **Concavity & Inflection Points**: Points where f''(x) = 0$.

  1. Point Plotting.

• Detailed Curve Tracing Examples:

  • Example 6.1 (Cissoid of Diocles): $y^2(2a - x) = x^3$ ($a > 0$)

  • Symmetric about X-axis. Cusp at origin ($2ay^2 = 0 \implies y = 0, 0$).

  • Vertical asymptote $x = 2a$. Region of absence: $x < 0$ and x > 2a$.\n    \n * **Example 6.2 (Strophoid)**: y^2(a - x) = x^2(a + x)$($a > 0)\n * Symmetric about X-axis. Node at origin (ay^2 - ax^2 = 0 \implies y = \pm x).\n * Intercepts (0, 0)$,$(-a, 0)$. Vertical asymptote$x = a$. Region of absence$x > a$or$x < -a$.

  • Vertical tangent at $(-a, 0)$.     

    

  • Example 6.3: $y = \frac{x^2}{1 - x^2}$

  • Symmetric about Y-axis. Tangent at origin $y = 0$. Asymptotes $y = -1$, $x = 1$, x = -1$.\n * **Example 6.4**: y^2(x - a) = x^2(x + a)$($a > 0)\n * Symmetric about X-axis. Origin is an isolated point (ay^2 + ax^2 = 0 \implies y = \pm ix).\n * Intercepts (0, 0)$,$(-a, 0)$. Vertical asymptote$x = a$. Oblique asymptotes$y = x + a$,$y = -x - a$.

  • Region of absence: $-a < x < a$ (except origin).

  • Example 6.5: $y = (x^2 - x - 6)(x - 7) = (x + 2)(x - 3)(x - 7)$

  • Intercepts $(-2, 0)$, $(3, 0)$, $(7, 0)$, $(0, 42)$. Stationary points at $x_1 = 0.063$ (Max: $42.0314$) and $x_2 = 5.27$ (Min: $-28.152$).

  • Example 6.6: $y = \frac{x^2 - 3x}{x - 1}$

  • Intercepts $(0, 0)$, $(3, 0)$. Tangent at origin $y = 3x$. Vertical asymptote $x = 1$. Oblique asymptote y = x - 2$.\n * **Example 6.7**: a^2 y^2 = x^2(2a - x)(x - a)\n * Symmetric about X-axis. Isolated point at origin. Curve exists only in a < x < 2a forming a closed loop.\n * Maximum at x = 1.64a$($y = 0.7872a$). Vertical tangents at$x = a$and$x = 2a$.

  • Example 6.8 (Folium of Descartes): $x^3 + y^3 = 9xy$

  • Symmetric about $y = x$ (intersects at $(4.5, 4.5)$). Node at origin ($x = 0, y = 0$).

  • Oblique asymptote x + y + 3 = 0$.\n\n\n# Curve Tracing in Polar Coordinates\n\n* **Polar Coordinate System**:\n * Position represented as (r, \theta)$, where$r$is distance from pole (origin) and$\theta is angle from polar axis (positive X-axis).\n\n* **Three Standard Polar Symmetry Tests**:\n 1. **Symmetry w.r.t. line \theta = \pi/2$(Y-axis)**: Equation remains unchanged when$(r, \theta)$is replaced by$(-r, -\theta)$or$(r, \pi - \theta).\n 2. **Symmetry w.r.t. Polar Axis (X-axis)**: Equation remains unchanged when (r, \theta)$is replaced by$(r, -\theta)$or$(-r, \pi - \theta).\n 3. **Symmetry w.r.t. Pole (Origin)**: Equation remains unchanged when (r, \theta)$is replaced by$(-r, \theta)$or$(r, \theta + \pi).\n\n* **Worked Examples**:\n * **Example 6.9 (Archimedean Spiral)**: r = \theta$on$[0, 2\pi]\n * As \theta$increases, distance$r$increases proportionally, forming a spiral originating from$(0, 0).\n * Key coordinates: (0, 0)$,$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$,$(\pi, \pi)$,$\left(\frac{3\pi}{2}, \frac{3\pi}{2}\right)$,$(2\pi, 2\pi).\n * **Example 6.10 (Cardioid)**: r = \cos(\theta) + 1$on$[0, 2\pi]\n * Symmetric about polar axis. Points: (2, 0)$,$\left(1.707, \frac{\pi}{4}\right)$,$\left(1, \frac{\pi}{2}\right)$,$\left(0.292, \frac{3\pi}{4}\right)$,$(0, \pi)$,$\left(1, \frac{3\pi}{2}\right)$,$(2, 2\pi).\n * **Example 6.11 (Limaçon with Inner Loop)**: r = 2\sin(\theta) - 1$on$[0, 2\pi]\n * Passes through \theta = 0 \implies r = -1 (plotted in Quadrant 3).\n * r = 0$at$\theta = \frac{\pi}{6}$and$\theta = \frac{5\pi}{6}.\n * Max r = 1$at$\theta = \frac{\pi}{2}$. Min$r = -3$at$\theta = \frac{3\pi}{2} (points to positive Y-axis due to negative radius).\n\n\n# Curve Tracing in Parametric Coordinates\n\n* **Parametric Representation**:\n * Coordinates defined by parameter t$:$x = f(t)$,$y = g(t)$for$t \in D\n\n* **Asymptotes in Parametric Form**:\n * **Horizontal Asymptote (y = k$)**: If$f(t) \rightarrow \infty$while$g(t) \rightarrow k$as$t \rightarrow t_0$.

  • Vertical Asymptote ($x = k$): If $f(t) \rightarrow k$ while $g(t) \rightarrow \infty$ as t \rightarrow t_0$.\n * **Oblique Asymptote (y = mx + c$)**: If both$f(t), g(t) \rightarrow \pm\infty$as$t \rightarrow t_0, then:\n    m = \lim_{t \rightarrow t_0} \frac{g(t)}{f(t)}, \quad c = \lim_{t \rightarrow t_0} [g(t) - m f(t)]\n\n* **Example 6.12 (Parametric Folium of Descartes)**:\n * Parametric equations: x = \frac{3at}{1 + t^3}$,$y = \frac{3at^2}{1 + t^3}$($a > 0)\n * As t \rightarrow -1^{-}$,$x \rightarrow -\infty$,$y \rightarrow \infty$. As$t \rightarrow -1^{+}$,$x \rightarrow \infty$,$y \rightarrow -\infty$.

  • Slope: $m = \lim_{t \rightarrow -1} \frac{3at^2 / (1 + t^3)}{3at / (1 + t^3)} = \lim_{t \rightarrow -1} t = -1$

  • Intercept: $c = \lim_{t \rightarrow -1} \left[\frac{3at^2}{1 + t^3} + \frac{3at}{1 + t^3}\right] = \lim_{t \rightarrow -1} \frac{3at(t + 1)}{(t + 1)(t^2 - t + 1)} = \frac{-3a}{3} = -a$

  • Inclined asymptote line: $y = -(x + a) \iff x + y + a = 0$

• Practice Homework Problems:

  1. $y = \frac{x^2}{1 - x^2}$

  2. $y = x^3 - 12x - 16$

  3. $x^3 + y^3 = 16xy$

  4. $a^2 y^2 = x^2(a^2 - x^2)$

  5. $y^2(x - a) = x^2(a + x)$

  6. $r = 2 + 2\cos(\theta)$

  7. $r^2 = a^2 \sin(2\theta)$

  8. $x = 2(\theta + \sin(\theta))$, $y = 2(1 + \cos(\theta))$