Chapter 12: Limits and Derivatives Calculus is the key to applying mathematics to explain the course of Nature. - WHITEHEAD 12.1 Introduction Cal

Chapter 12: Limits and Derivatives 1

Calculus is the key to applying mathematics to explain the course of Nature. - WHITEHEAD

12.1 Introduction 1

Calculus is the branch of mathematics that studies how function values change as their domain points change. This chapter introduces:

• An intuitive idea of the derivative.

• A basic definition of limits and their algebraic properties.

• The formal definition of the derivative and its algebraic properties.

• Derivatives of standard functions.

12.2 Intuitive Idea of Derivatives 1-4

This section explores the concept of instantaneous rate of change using the example of a falling body.

Falling Body Example:A body dropped from a tall cliff covers a distance $s = 4.9 t^2$ meters in $t$ seconds. To find the velocity at $t=2$ seconds, we calculate the average velocity over progressively smaller time intervals ending at $t=2$.

Average Velocity Calculation:Average velocity between $t_1$ and $t_2$ is $\frac{\text{Distance traveled}}{\text{Time interval}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$.

• Average velocity between $t=0$ and $t=2$ seconds:$\frac{(4.9(2)^2 - 4.9(0)^2) \text{ m}}{(2-0) \text{ s}} = \frac{19.6}{2} = 9.8 \text{ m/s}$

• Average velocity between $t=1$ and $t=2$ seconds:$\frac{(4.9(2)^2 - 4.9(1)^2) \text{ m}}{(2-1) \text{ s}} = \frac{19.6 - 4.9}{1} = 14.7 \text{ m/s}$

Table 12.1: Distance traveled at various times. 2

Table 12.2: Average velocity between $t_1$ and $t=2$ seconds. 2

Table 12.3: Average velocity between $t=2$ seconds and $t_2$ seconds. 3

Observation: As the time interval shrinks, the average velocity approaches a specific value. This value represents the instantaneous velocity at $t=2$ seconds, which is between $19.551 \text{ m/s}$ and $19.649 \text{ m/s}$. This instantaneous velocity is the derivative of the distance function at $t=2$.

Graphical Interpretation:

The derivative at a point is the slope of the tangent line to the curve at that point.

Fig 12.1: Distance $s$ vs. Time $t$ showing secant line slopes approaching tangent slope. 4

12.3 Limits 4-12

The limiting process is crucial for understanding derivatives. A limit describes the value a function approaches as its input approaches a certain value.

• Definition: If a function $f(x)$ approaches $l$ as $x$ approaches $a$, then $l$ is the limit of $f(x)$ as $x$ tends to $a$. Symbolically: $\lim _{x \rightarrow a} f(x)=l$.

• The limit is concerned with the behavior of the function near a point, not necessarily at the point itself.

Illustrative Examples:

• $\lim _{x \rightarrow 0} x^2 = 0$ (as $x$ gets close to 0, $x^2$ gets close to 0). 4

• For $g(x)=|x|$ where $x \neq 0$, $\lim _{x \rightarrow 0} g(x)=0$. 4

• For $h(x)=\frac{x^2-4}{x-2}$ where $x \neq 2$, as $x$ approaches 2, $h(x)$ approaches 4. $\lim _{x \rightarrow 2} h(x)=4$. 4

Left-Hand and Right-Hand Limits 5

There are two ways $x$ can approach $a$: from the left (values less than $a$) or from the right (values greater than $a$).

• Left-hand limit: The value $f(x)$ approaches as $x$ approaches $a$ from the left. Denoted as $\lim _{x \rightarrow a^{-}} f(x)$.

• Right-hand limit: The value $f(x)$ approaches as $x$ approaches $a$ from the right. Denoted as $\lim _{x \rightarrow a^{+}} f(x)$.

Condition for Limit Existence:The limit $\lim _{x \rightarrow a} f(x)$ exists if and only if the left-hand limit and the right-hand limit exist and are equal: $\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)$ If they are equal, then $\lim _{x \rightarrow a} f(x)$ is this common value.

Example of Non-Existent Limit:Consider the function: $f(x)= \begin{cases} 1, & x \leq 0 \\ 2, & x>0 \end{cases}$

• Left-hand limit at $x=0$: $\lim _{x \rightarrow 0^{-}} f(x)=1$

• Right-hand limit at $x=0$: $\lim _{x \rightarrow 0^{+}} f(x)=2$ Since $1 \neq 2$, $\lim _{x \rightarrow 0} f(x)$ does not exist. 5

Fig 12.3: Graph of a piecewise function showing differing left and right limits at $x=0$. 5

Illustrative Examples:

• Illustration 1: For $f(x)=x+10$, $\lim _{x \rightarrow 5^{-}} f(x)=15$ and $\lim _{x \rightarrow 5^{+}} f(x)=15$. Thus, $\lim _{x \rightarrow 5} f(x)=15$. 6

• Illustration 2: For $f(x)=x^3$, $\lim _{x \rightarrow 1^{-}} f(x)=1$ and $\lim _{x \rightarrow 1^{+}} f(x)=1$. Thus, $\lim _{x \rightarrow 1} f(x)=1$. 7

• Illustration 3: For $f(x)=3x$, $\lim _{x \rightarrow 2^{-}} f(x)=6$ and $\lim _{x \rightarrow 2^{+}} f(x)=6$. Thus, $\lim _{x \rightarrow 2} f(x)=6$. 7

• Illustration 4: For $f(x)=3$ (constant function), $\lim _{x \rightarrow 2} f(x)=3$. 8

• Illustration 5: For $f(x)=x^2+x$, $\lim _{x \rightarrow 1^{-}} f(x)=2$ and $\lim _{x \rightarrow 1^{+}} f(x)=2$. Thus, $\lim _{x \rightarrow 1} f(x)=2$. 8

• Illustration 6: For $f(x)=\sin x$, $\lim _{x \rightarrow \frac{\pi}{2}} \sin x=1$. 9

• Illustration 7: For $f(x)=x+\cos x$, $\lim _{x \rightarrow 0} f(x)=1$. 9

• Illustration 8: For $f(x)=\frac{1}{x^2}$ where x>0, as $x \rightarrow 0$, $f(x)$ becomes arbitrarily large. We write $\lim _{x \rightarrow 0} f(x)=+\infty$. 10

• Illustration 9: For $f(x)=\begin{cases} x-2, & x<0 \\ 0, & x=0 \\ x+2, & x>0 \end{cases}$, $\lim _{x \rightarrow 0^{-}} f(x)=-2$ and $\lim _{x \rightarrow 0^{+}} f(x)=2$. The limit does not exist. 10

• Illustration 10: For $f(x)=\begin{cases} x+2, & x \neq 1 \\ 0, & x=1 \end{cases}$, $\lim _{x \rightarrow 1^{-}} f(x)=3$ and $\lim _{x \rightarrow 1^{+}} f(x)=3$. Thus, $\lim _{x \rightarrow 1} f(x)=3$. Note that $\lim _{x \rightarrow 1} f(x) \neq f(1)$. 11

12.3.1 Algebra of Limits 12-17

If $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ exist, then:

• Sum Rule: $\lim _{x \rightarrow a}[f(x)+g(x)]=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)$

• Difference Rule: $\lim _{x \rightarrow a}[f(x)-g(x)]=\lim _{x \rightarrow a} f(x)-\lim _{x \rightarrow a} g(x)$

• Product Rule: $\lim _{x \rightarrow a}[f(x) \cdot g(x)]=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)$

• Quotient Rule: $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}$ (provided $\lim _{x \rightarrow a} g(x) \neq 0$)

• Constant Multiple Rule: $\lim _{x \rightarrow a}[\lambda \cdot f(x)]=\lambda \cdot \lim _{x \rightarrow a} f(x)$

12.3.2 Limits of Polynomial and Rational Functions 12-16

• Polynomial Function: $f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$. For a polynomial function, $\lim _{x \rightarrow a} f(x) = f(a)$. This is because: $\lim _{x \rightarrow a} x^n = a^n$ and the limit rules for sums and constant multiples.

• Rational Function: $f(x) = \frac{g(x)}{h(x)}$, where $g(x)$ and $h(x)$ are polynomials and $h(x) \neq 0$.

  • If $h(a) \neq 0$, then $\lim _{x \rightarrow a} f(x) = \frac{g(a)}{h(a)}$.

  • If $h(a) = 0$ and $g(a) \neq 0$, the limit does not exist (or tends to $\pm \infty$).

  • If $h(a) = 0$ and $g(a) = 0$, the limit is of the indeterminate form $\frac{0}{0}$. Algebraic manipulation (factoring and canceling) is often needed.

Example 1: Find the limits: 14

• (i) $\lim _{x \rightarrow 1}\left[x^{3}-x^{2}+1\right] = 1^3 - 1^2 + 1 = 1$

• (ii) $\lim _{x \rightarrow 3}[x(x+1)] = 3(3+1) = 12$

• (iii) $\lim _{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right] = 1-1+1-1+\dots+1 = 1$

Example 2: Find the limits: 14-16

• (i) $\lim _{x \rightarrow 1}\left[\frac{x^{2}+1}{x+100}\right]=\frac{1^2+1}{1+100}=\frac{2}{101}$

• (ii) $\lim _{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}\right]$ This is of the form $\frac{0}{0}$. Factorizing: $\lim _{x \rightarrow 2} \frac{x(x-2)^2}{(x+2)(x-2)} = \lim _{x \rightarrow 2} \frac{x(x-2)}{(x+2)} = \frac{2(0)}{4} = 0$

• (iii) $\lim _{x \rightarrow 2}\left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]$ This is of the form $\frac{0}{0}$. Factorizing: $\lim _{x \rightarrow 2} \frac{(x-2)(x+2)}{x(x-2)^2} = \lim _{x \rightarrow 2} \frac{x+2}{x(x-2)} = \frac{4}{0}$ The limit does not exist.

• (iv) $\lim _{x \rightarrow 2}\left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$ This is of the form $\frac{0}{0}$. Factorizing: $\lim _{x \rightarrow 2} \frac{x^2(x-2)}{(x-2)(x-3)} = \lim _{x \rightarrow 2} \frac{x^2}{x-3} = \frac{2^2}{2-3} = \frac{4}{-1} = -4$

• (v) $\lim _{x \rightarrow 1}\left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]$ Combine fractions: $\frac{x^{2}-4 x+3}{x(x-1)(x-2)}$ This is of the form $\frac{0}{0}$ at $x=1$. $\lim _{x \rightarrow 1} \frac{(x-1)(x-3)}{x(x-1)(x-2)} = \lim _{x \rightarrow 1} \frac{x-3}{x(x-2)} = \frac{1-3}{1(1-2)} = \frac{-2}{-1} = 2$

Important Limit Theorem: For a positive integer $n$: $\lim _{x \rightarrow a} \frac{x^n - a^n}{x-a} = n a^{n-1}$ This holds even if $n$ is any rational number and a>0. 16

Example 3: Evaluate: 17

• (i) $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$ $\lim _{x \rightarrow 1} \frac{\frac{x^{15}-1}{x-1}}{\frac{x^{10}-1}{x-1}} = \frac{15(1)^{14}}{10(1)^9} = \frac{15}{10} = \frac{3}{2}$

• (ii) $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}$ Let $y=1+x$. As $x \rightarrow 0$, $y \rightarrow 1$. $\lim _{y \rightarrow 1} \frac{y^{1/2}-1^{1/2}}{y-1} = \frac{1}{2}(1)^{1/2-1} = \frac{1}{2}$

12.4 Limits of Trigonometric Functions 18-21

Comparison Theorem: If $f(x) \leq g(x)$ for all $x$ near $a$, and $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ exist, then $\lim _{x \rightarrow a} f(x) \leq \lim _{x \rightarrow a} g(x)$. 18

Sandwich Theorem (Squeeze Theorem): If $f(x) \leq g(x) \leq h(x)$ for all $x$ near $a$, and $\lim _{x \rightarrow a} f(x) = l = \lim _{x \rightarrow a} h(x)$, then $\lim _{x \rightarrow a} g(x) = l$. 18

Key Inequality: For 0 < |x| < \frac{\pi}{2}: \cos x < \frac{\sin x}{x} < 1 This inequality is proven geometrically using areas of triangles and sectors in a unit circle. 19

Important Trigonometric Limits:

• (i) $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

• (ii) $\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=0$

Proof of (i): Follows directly from the Sandwich Theorem, using the inequality \cos x < \frac{\sin x}{x} < 1 and knowing $\lim _{x \rightarrow 0} \cos x = 1$. 20Proof of (ii): Using the identity $1-\cos x = 2 \sin^2(\frac{x}{2})$: $\lim _{x \rightarrow 0} \frac{2 \sin^2(\frac{x}{2})}{x} = \lim _{x \rightarrow 0} \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \cdot \sin(\frac{x}{2}) = 1 \cdot 0 = 0$

Example 4: Evaluate: 20

• (i) $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}$ $\lim _{x \rightarrow 0} \left( \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot 2 \right) = 1 \cdot 1 \cdot 2 = 2$

• (ii) $\lim _{x \rightarrow 0} \frac{\tan x}{x}$ $\lim _{x \rightarrow 0} \frac{\sin x}{x \cos x} = \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \frac{1}{\cos x} = 1 \cdot 1 = 1$

General Strategy for Limits of the form $\frac{f(x)}{g(x)}$:

1. Evaluate $f(a)$ and $g(a)$.

2. If both are 0, try to factor out the term causing the zero (e.g., $(x-a)$).

3. Rewrite the expression as $\frac{p(x)}{q(x)}$ where $q(x) \neq 0$.

4. Evaluate $\frac{p(a)}{q(a)}$.

Exercise 12.1 21-23

Evaluate the following limits:

1. $\lim _{x \rightarrow 3} (x+3) = 6$

2. $\lim _{x \rightarrow \pi}\left(x-\frac{22}{7}\right) = \pi - \frac{22}{7}$

3. $\lim _{r \rightarrow 1} \pi r^{2} = \pi (1)^2 = \pi$

4. $\lim _{x \rightarrow 4} \frac{4 x+3}{x-2} = \frac{4(4)+3}{4-2} = \frac{19}{2}$

5. $\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1} = \frac{(-1)^{10}+(-1)^{5}+1}{-1-1} = \frac{1-1+1}{-2} = -\frac{1}{2}$

6. $\lim _{x \rightarrow 0} \frac{(x+1)^{5}-1}{x} = 5(1)^{4} = 5$ (Using Theorem: $\lim _{x \rightarrow a} \frac{x^n - a^n}{x-a}=n a^{n-1}$)

7. $\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}$ (Form $\frac{0}{0}$) $\lim _{x \rightarrow 2} \frac{(3x+5)(x-2)}{(x-2)(x+2)} = \lim _{x \rightarrow 2} \frac{3x+5}{x+2} = \frac{3(2)+5}{2+2} = \frac{11}{4}$

8. $\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}$ (Form $\frac{0}{0}$) $\lim _{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(2x+1)(x-3)} = \lim _{x \rightarrow 3} \frac{(x+3)(x^2+9)}{2x+1} = \frac{(3+3)(3^2+9)}{2(3)+1} = \frac{6 \cdot 18}{7} = \frac{108}{7}$

9. $\lim _{x \rightarrow 0} \frac{a x+b}{c x+1} = \frac{a(0)+b}{c(0)+1} = b$

10. $\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{6}-1}$ (Form $\frac{0}{0}$) $\lim _{z \rightarrow 1} \frac{\frac{z^{1/3}-1}{z-1}}{\frac{z^6-1}{z-1}} = \frac{\frac{1}{3}(1)^{1/3-1}}{6(1)^{6-1}} = \frac{1/3}{6} = \frac{1}{18}$

11. $\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = \frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a} = \frac{a+b+c}{a+b+c} = 1$ (since $a+b+c \neq 0$)

12. $\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$ (Form $\frac{0}{0}$) $\lim _{x \rightarrow-2} \frac{\frac{2+x}{2x}}{x+2} = \lim _{x \rightarrow-2} \frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}$

13. $\lim _{x \rightarrow 0} \frac{\sin a x}{b x} = \lim _{x \rightarrow 0} \frac{\sin a x}{a x} \cdot \frac{a}{b} = 1 \cdot \frac{a}{b} = \frac{a}{b}$

14. $\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} = \lim _{x \rightarrow 0} \frac{\frac{\sin a x}{a x}}{\frac{\sin b x}{b x}} \cdot \frac{a}{b} = \frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}$

15. $\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$ Let $y=\pi-x$. As $x \rightarrow \pi$, $y \rightarrow 0$. $\lim _{y \rightarrow 0} \frac{\sin y}{\pi y} = \frac{1}{\pi} \lim _{y \rightarrow 0} \frac{\sin y}{y} = \frac{1}{\pi} \cdot 1 = \frac{1}{\pi}$

16. $\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x} = \frac{\cos 0}{\pi-0} = \frac{1}{\pi}$

17. $\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$ (Form $\frac{0}{0}$) Use $1-\cos 2x = 2\sin^2 x$ and $1-\cos x = 2\sin^2(x/2)$. $\lim _{x \rightarrow 0} \frac{-(1-\cos 2 x)}{-(1-\cos x)} = \lim _{x \rightarrow 0} \frac{2 \sin^2 x}{2 \sin^2(x/2)} = \lim _{x \rightarrow 0} \left( \frac{\sin x}{\sin(x/2)} \right)^2$ $= \lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{x}{x/2} \right)^2 = \left( 1 \cdot 1 \cdot 2 \right)^2 = 4$

18. $\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x} = \lim _{x \rightarrow 0} \frac{x(a+\cos x)}{b \sin x} = \lim _{x \rightarrow 0} \frac{a+\cos x}{b \frac{\sin x}{x}} = \frac{a+\cos 0}{b \cdot 1} = \frac{a+1}{b}$

19. $\lim _{x \rightarrow 0} x \sec x = \lim _{x \rightarrow 0} \frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0$

20. $\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x}$ (Form $\frac{0}{0}$) Divide numerator and denominator by $x$: $\lim _{x \rightarrow 0} \frac{\frac{\sin a x}{x}+b}{a+\frac{\sin b x}{x}} = \lim _{x \rightarrow 0} \frac{a \frac{\sin a x}{a x}+b}{a+b \frac{\sin b x}{b x}} = \frac{a(1)+b}{a+b(1)} = \frac{a+b}{a+b} = 1$

21. $\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x) = \lim _{x \rightarrow 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) = \lim _{x \rightarrow 0} \frac{1-\cos x}{\sin x}$ (Form $\frac{0}{0}$) $= \lim _{x \rightarrow 0} \frac{2 \sin^2(x/2)}{2 \sin(x/2) \cos(x/2)} = \lim _{x \rightarrow 0} \tan(x/2) = \tan(0) = 0$

22. $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}$ Let $y=x-\frac{\pi}{2}$. Then $x=y+\frac{\pi}{2}$, and $2x = 2y+\pi$. As $x \rightarrow \frac{\pi}{2}$, $y \rightarrow 0$. $\lim _{y \rightarrow 0} \frac{\tan(2y+\pi)}{y} = \lim _{y \rightarrow 0} \frac{\tan(2y)}{y} = \lim _{y \rightarrow 0} \frac{\sin(2y)}{y \cos(2y)} = \lim _{y \rightarrow 0} \frac{\sin(2y)}{2y} \cdot \frac{2}{\cos(2y)} = 1 \cdot \frac{2}{1} = 2$

23. $f(x)=\begin{cases} 2 x+3, &amp; x \leq 0 \\ 3(x+1), &amp; x&gt;0 \end{cases}$

    • $\lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} (2x+3) = 3$

    • $\lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} 3(x+1) = 3$ Therefore, $\lim _{x \rightarrow 0} f(x) = 3$.

    • For $\lim _{x \rightarrow 1} f(x)$, since x>0 for $x$ near 1, we use $f(x)=3(x+1)$.

    • $\lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} 3(x+1) = 3(1+1) = 6$.

24. $f(x)=\begin{cases} x^{2}-1, &amp; x \leq 1 \\ -x^{2}-1, &amp; x&gt;1 \end{cases}$

    • $\lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (x^2-1) = 1^2-1 = 0$

    • $\lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (-x^2-1) = -(1)^2-1 = -2$ The limit does not exist.

25. $f(x)=\begin{cases} \frac{|x|}{x}, &amp; x \neq 0 \\ 0, &amp; x=0 \end{cases}$

    • For x>0, $f(x)=\frac{x}{x}=1$. So $\lim _{x \rightarrow 0^{+}} f(x)=1$.

    • For x<0, $f(x)=\frac{-x}{x}=-1$. So $\lim _{x \rightarrow 0^{-}} f(x)=-1$. The limit does not exist.

26. $f(x)=\begin{cases} \frac{x}{|x|}, &amp; x \neq 0 \\ 0, &amp; x=0 \end{cases}$

    • For x>0, $f(x)=\frac{x}{x}=1$. So $\lim _{x \rightarrow 0^{+}} f(x)=1$.

    • For x<0, $f(x)=\frac{x}{-x}=-1$. So $\lim _{x \rightarrow 0^{-}} f(x)=-1$. The limit does not exist.

27. $\lim _{x \rightarrow 5} f(x)=|x|-5$. Since $|x|-5$ is continuous at $x=5$, the limit is $|5|-5 = 5-5=0$.

28. $f(x)= \begin{cases} a+b x, &amp; x&lt;1 \\ 4, &amp; x=1 \\ b-a x, &amp; x&gt;1 \end{cases}$ If $\lim _{x \rightarrow 1} f(x)=f(1)$, then the left-hand limit must equal the right-hand limit, and both must equal $f(1)=4$.

    • $\lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (a+bx) = a+b = 4$

    • $\lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (b-ax) = b-a = 4$ We have a system of equations: $a+b = 4$ $b-a = 4$ Adding the two equations: $2b = 8 \implies b=4$. Substituting $b=4$ into $a+b=4$: $a+4=4 \implies a=0$. So, $a=0, b=4$.

29. $f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)$

    • $\lim _{x \rightarrow a_{1}} f(x)$: Since $f(x)$ is a polynomial, the limit is $f(a_1)$. If $a_1$ is one of the roots $a_i$, then $f(a_1)=0$. If $a_1$ is not among $a_2, \dots, a_n$, then $f(a_1) = (a_1-a_1)(a_1-a_2)\dots(a_1-a_n) = 0$. So, $\lim _{x \rightarrow a_{1}} f(x) = 0$.

    • For $a \neq a_{1}, a_{2}, \ldots, a_{n}$, $\lim _{x \rightarrow a} f(x) = f(a) = (a-a_{1})(a-a_{2}) \ldots(a-a_{n})$.

30. $f(x)=\left\{\begin{array}{ll} |x|+1, &amp; x&lt;0 \\ 0, &amp; x=0 \\ |x|-1, &amp; x&gt;0 \end{array}\right.$

    • For x<0, $f(x)=-x+1$. $\lim _{x \rightarrow 0^{-}} f(x) = -0+1 = 1$.

    • For x>0, $f(x)=x-1$. $\lim _{x \rightarrow 0^{+}} f(x) = 0-1 = -1$. The limit at $x=0$ does not exist.

    • For a>0, $\lim _{x \rightarrow a} f(x) = \lim _{x \rightarrow a} (x-1) = a-1$. This exists for all a>0.

    • For a<0, $\lim _{x \rightarrow a} f(x) = \lim _{x \rightarrow a} (-x+1) = -a+1$. This exists for all a<0. The limit exists for all $a \neq 0$.

31. Given $\lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi$. Let $L = \lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}$. Since $x^2-1 \to 0$ as $x \to 1$, for the limit $L$ to exist (and be finite, $\pi$), the numerator must also approach 0. So, $\lim _{x \rightarrow 1} (f(x)-2) = 0$, which means $\lim _{x \rightarrow 1} f(x) = 2$.

32. $f(x)=\left\{\begin{array}{ll}m x^{2}+n, &amp; x&lt;0 \\ n x+m, &amp; 0 \leq x \leq 1\end{array}\right.$.

    • For $\lim _{x \rightarrow 0} f(x)$ to exist: $\lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} (mx^2+n) = n$. $\lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} (nx+m) = m$. So, we need $n=m$.

    • For $\lim _{x \rightarrow 1} f(x)$ to exist: $\lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (nx+m) = n(1)+m = n+m$. $\lim _{x \rightarrow 1^{+}} f(x)$ is not defined by the given function, as the second case is $0 \leq x \leq 1$. Assuming the domain is extended or this is a typo and it should be x>1 for some other function. If we only consider the defined intervals, we only need to ensure the limit at 0 exists. If we assume the question implies the function is defined piecewise for x<0 and $0 \le x \le 1$, then we only need to worry about the limit at $x=0$. For the limit at $x=1$ to exist from the left, $\lim_{x \to 1^-} f(x) = n+m$. If the question implied continuity at $x=1$, then the function definition would need to extend beyond $x=1$. Assuming the question only asks for the limits at $x=0$ and $x=1$ to exist based on the provided definition: Limit at $x=0$ exists if $m=n$. Limit at $x=1$ from the left exists if $n+m$ is defined (which it is, as $n,m$ are integers). If the question meant to ask for continuity at $x=1$, we would need a definition for x>1. If the question implies the limit at $x=1$ exists within the defined domain, it means the left-hand limit exists. Let's re-read: "For what integers m and n does both $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$ exist?" This implies we need the limit at $x=1$ to exist. Based on the definition, the function is only defined up to $x=1$. This usually means we only consider the limit from the left at $x=1$. So, for $\lim _{x \rightarrow 0} f(x)$ to exist, $m=n$. For $\lim _{x \rightarrow 1} f(x)$ to exist (as a left-hand limit), it must be $n+m$. This limit exists for any integers $m, n$. Thus, the condition is $m=n$.