Comprehensive Notes on Limits, Trigonometric Identities, and the Squeeze Theorem

Fundamental Trigonometric Identities and Limits

• Core Pythagorean Identity:$\sin^2(\theta) + \cos^2(\theta) = 1$

• Double-Angle Identity for Cosine:$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 1 - 2\sin^2(\theta)$

• Fundamental Cosine Limit:$\lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\theta} = 0$

• Analysis of Derivation Attempts for Double-Angle Cosine Limit:

  • Evaluating the target limit expression:         $\lim_{\theta \to 0} \frac{\cos(2\theta) - \cos(\theta)}{\theta}$
  • Attempt 1 (Double-Angle Expansion): Substituting $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$$\lim_{\theta \to 0} \frac{\cos^2(\theta) - \sin^2(\theta) - \cos(\theta)}{\theta}$
  • Attempt 2 (Factoring Difference of Squares): Factoring $\cos^2(\theta) - \sin^2(\theta)$$\lim_{\theta \to 0} \frac{(\cos(\theta) + \sin(\theta))(\cos(\theta) - \sin(\theta)) - \cos(\theta)}{\theta}$Note: This approach does not resolve the indeterminate form.
  • Attempt 3 (Pythagorean Substitution): Substituting $\cos^2(\theta) = 1 - \sin^2(\theta)$$\lim_{\theta \to 0} \frac{1 - \sin^2(\theta) - \sin^2(\theta) - \cos(\theta)}{\theta} = \lim_{\theta \to 0} \frac{1 - 2\sin^2(\theta) - \cos(\theta)}{\theta}$Note: Simple substitution without separating known limit structures leaves indeterminate components.

The Squeeze Theorem and Special Trigonometric Limits

• Definition of the Squeeze Theorem (Sandwich Theorem):

  • Condition: If $l(x) \le f(x) \le u(x)$ for all $x$ in an open interval around $a$ (where $u(x)$ represents the upper bound function and $l(x)$ represents the lower bound function), and:         $\lim_{x \to a} l(x) = L \quad \text{and} \quad \lim_{x \to a} u(x) = L$
  • Conclusion:         $\lim_{x \to a} f(x) = L$
  • Concept: The function $f(x)$ is trapped or squeezed between $l(x)$ and $u(x)$, forcing its limit at point $a$ to equal $L$.
  • Applicability note: Applies directly when evaluating bounded limits as limits approach zero.

• Fundamental Standard Trigonometric Limits:$\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$$\lim_{\theta \to 0} \frac{\theta}{\sin(\theta)} = 1$$\lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\theta} = 0$

• Argument-Matching Rule:

  • To apply $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$, the argument of the sine function must exactly match the expression in the denominator.

• Worked Examples of Trigonometric Limits:

  • Example 1: Evaluate $\lim_{h \to 0} \frac{\sin(4h)}{h}$
    • Multiply numerator and denominator by $4$ to align arguments:             $\lim_{h \to 0} \frac{4\sin(4h)}{4h} = 4 \cdot \lim_{h \to 0} \frac{\sin(4h)}{4h} = 4 \cdot 1 = 4$
  • Example 2: Evaluate $\lim_{x \to 0} \frac{\tan(3x)}{\tan(2x)}$
    • Express tangent in terms of sine and cosine:             $\frac{\tan(3x)}{\tan(2x)} = \frac{\frac{\sin(3x)}{\cos(3x)}}{\frac{\sin(2x)}{\cos(2x)}} = \frac{\sin(3x)}{\cos(3x)} \cdot \frac{\cos(2x)}{\sin(2x)}$
    • Regroup factors:             $\left(\frac{\sin(3x)}{\sin(2x)}\right) \cdot \left(\frac{\cos(2x)}{\cos(3x)}\right)$
    • Multiply and divide by arguments $3x$ and $2x$:             $\left(\frac{\sin(3x)}{3x}\right) \cdot \left(\frac{2x}{\sin(2x)}\right) \cdot \frac{3x}{2x} \cdot \frac{\cos(2x)}{\cos(3x)}$
    • Take limit as $x \to 0$:             $1 \cdot 1 \cdot \frac{3}{2} \cdot \frac{\cos(0)}{\cos(0)} = 1 \cdot 1 \cdot \frac{3}{2} \cdot 1 = \frac{3}{2}$
  • Example 3: Evaluate $\lim_{x \to 0} \frac{6\sin(x)}{2x}$
    • Factor out constants:             $\lim_{x \to 0} \frac{6}{2} \cdot \left(\frac{\sin(x)}{x}\right) = 3 \cdot 1 = 3$
    • Variation $\lim_{x \to 0} \frac{\sin(x)}{2x}$:             $\lim_{x \to 0} \frac{1}{2} \left(\frac{\sin(x)}{x}\right) = \frac{1}{2}(1) = \frac{1}{2}$
  • Example 4: Evaluate $\lim_{x \to 0} \frac{\sin(2x)}{x}$
    • Multiply numerator and denominator by $2$:             $\lim_{x \to 0} 2 \cdot \left(\frac{\sin(2x)}{2x}\right) = 2(1) = 2$
  • Example 5: Evaluate $\lim_{t \to \frac{\pi}{4}} \frac{\sin(t)}{t}$
    • Apply direct substitution since $t \to \frac{\pi}{4} \neq 0$:             $\frac{\sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{4}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\pi}{4}} = \frac{\sqrt{2}}{2} \cdot \frac{4}{\pi} = \frac{4\sqrt{2}}{2\pi} = \frac{2\sqrt{2}}{\pi}$
  • Example 6: Evaluate $\lim_{\theta \to 0} \frac{\sin(7\theta)}{\sin(3\theta)}$
    • Multiply numerator and denominator by matching arguments:             $\lim_{\theta \to 0} \left(\frac{\sin(7\theta)}{7\theta}\right) \cdot \left(\frac{3\theta}{\sin(3\theta)}\right) \cdot \frac{7\theta}{3\theta} = 1 \cdot 1 \cdot \frac{7}{3} = \frac{7}{3}$
    • General Rule: Coefficients balance out such that $\lim_{\theta \to 0} \frac{\sin(a\theta)}{\sin(b\theta)} = \frac{a}{b}$.
  • Example 7 (Substitution Technique): Evaluate $\lim_{\theta \to 0} \frac{\sin(2\theta)}{7\theta}$
    • Let $u = 2\theta$, which implies $\theta = \frac{u}{2}$. As $\theta \to 0$, $u \to 0$
    • Substitute $u$ into the expression:             $\lim_{u \to 0} \frac{\sin(u)}{7\left(\frac{u}{2}\right)} = \lim_{u \to 0} \frac{2\sin(u)}{7u} = \frac{2}{7} \lim_{u \to 0} \frac{\sin(u)}{u} = \frac{2}{7}(1) = \frac{2}{7}$

Algebraic Limit Techniques and Indeterminate Forms

• Technique 1: Combining Fractions for Indeterminate Forms of Type $\left[\infty - \infty\right]$

  • Evaluate $\lim_{x \to 1} \left( \frac{1}{x-1} - \frac{2}{x^2-1} \right)$
  • Direct substitution yields $\frac{1}{0} - \frac{2}{0} = \infty - \infty$ (indeterminate form).
  • Factor denominator and determine common denominator:         $\frac{1}{x-1} - \frac{2}{(x-1)(x+1)} = \frac{x+1}{(x-1)(x+1)} - \frac{2}{(x-1)(x+1)} = \frac{x+1-2}{(x-1)(x+1)} = \frac{x-1}{(x-1)(x+1)}$
  • Cancel common factor $(x-1)$ for $x \neq 1$:         $\frac{1}{x+1}$
  • Evaluate limit as $x \to 1$:         $\lim_{x \to 1} \frac{1}{x+1} = \frac{1}{1+1} = \frac{1}{2}$

• Technique 2: Expanding Difference Quotients for Indeterminate Forms of Type $\left[\frac{0}{0}\right]$

  • Evaluate $\lim_{h \to 0} \frac{(h+a)^2 - a^2}{h}$
  • Direct substitution yields $\frac{(0+a)^2 - a^2}{0} = \frac{a^2 - a^2}{0} = \frac{0}{0}$ (indeterminate form).
  • Expand algebraic expression in numerator:         $(h+a)^2 - a^2 = h^2 + 2ah + a^2 - a^2 = h^2 + 2ah$
  • Factor out $h$:         $\frac{h(h + 2a)}{h}$
  • Cancel factor $h$ for $h \neq 0$:         $h + 2a$
  • Evaluate limit as $h \to 0$:         $\lim_{h \to 0} (h + 2a) = 0 + 2a = 2a$

• Technique 3: Trigonometric Identity Simplification

  • Evaluate $\lim_{x \to \frac{\pi}{4}} \frac{\sin(x) - \cos(x)}{\tan(x) - 1}$
  • Direct substitution yields $\frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{1 - 1} = \frac{0}{0}$ (indeterminate form).
  • Rewrite $\tan(x) = \frac{\sin(x)}{\cos(x)}$ in denominator:         $\tan(x) - 1 = \frac{\sin(x)}{\cos(x)} - 1 = \frac{\sin(x) - \cos(x)}{\cos(x)}$
  • Simplify overall fraction:         $\frac{\sin(x) - \cos(x)}{\frac{\sin(x) - \cos(x)}{\cos(x)}} = (\sin(x) - \cos(x)) \cdot \frac{\cos(x)}{\sin(x) - \cos(x)} = \cos(x)$
  • Evaluate limit as $x \to \frac{\pi}{4}$:         $\lim_{x \to \frac{\pi}{4}} \cos(x) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

• Technique 4: Factoring Quadratic Expressions

  • Evaluate $\lim_{t \to 2} \frac{2t^2 + 2t - 20}{2t - 4}$
  • Direct substitution yields $\frac{2(2)^2 + 2(2) - 20}{2(2) - 4} = \frac{16 + 4 - 20}{4 - 4} = \frac{0}{0}$ (indeterminate form).
  • Factor numerator and denominator:         $\frac{(2t + 5)(2t - 4)}{2t - 4}$
  • Cancel common factor $(2t - 4)$:         $\lim_{t \to 2} (2t + 5)$
  • Evaluate limit:         $2(2) + 5 = 4 + 5 = 9$

Limits at Infinity and Rational Functions

• Asymptotic Behavior of Elementary Functions:

  • Quadratic Function ($f(x) = x^2$):$\lim_{x \to \infty} x^2 = \infty \quad \text{and} \quad \lim_{x \to -\infty} x^2 = \infty$
  • Cubic Function ($f(x) = x^3$):$\lim_{x \to \infty} x^3 = \infty \quad \text{and} \quad \lim_{x \to -\infty} x^3 = -\infty$
  • Reciprocal Function ($f(x) = \frac{1}{x}$):$\lim_{x \to 0^+} \frac{1}{x} = \infty \quad \text{and} \quad \lim_{x \to 0^-} \frac{1}{x} = -\infty$$\lim_{x \to \infty} \frac{1}{x} = 0 \quad \text{and} \quad \lim_{x \to -\infty} \frac{1}{x} = 0$
  • Oscillatory Functions:$\lim_{x \to \infty} \sin(x) \quad \text{is undefined because the function oscillates perpetually between } -1 \text{ and } 1.$
  • General Reciprocal Power Rule:$\lim_{x \to \infty} \frac{n}{x^p} = 0 \quad \text{for any real constant } n \text{ and positive power } p > 0.$
  • Exponential Function ($f(x) = e^x$):$\lim_{x \to \infty} e^x = \infty \quad \text{and} \quad \lim_{x \to -\infty} e^x = 0$

• General Degree Rules for Rational Functions Limits at Infinity:

  • Rule 1 (Numerator degree > Denominator degree): The limit equals $\infty$ or $-\infty$.
  • Rule 2 (Denominator degree > Numerator degree): The limit equals $0$.
  • Rule 3 (Numerator degree = Denominator degree): The limit equals the ratio of the leading coefficients $\frac{a}{b}$.

• Method of Dividing by the Highest Order of Infinity:

  • Example 1 (Rational function with fractional exponents): Evaluate $\lim_{x \to \infty} \frac{3x^{7/2} + 7x^{-1/2}}{x^2 - x^{1/2}}$
    • Direct substitution yields $\left[\frac{\infty}{\infty}\right]$.
    • Divide numerator and denominator by highest power in denominator ($x^2$):             $\lim_{x \to \infty} \frac{\frac{3x^{7/2}}{x^2} + \frac{7x^{-1/2}}{x^2}}{\frac{x^2}{x^2} - \frac{x^{1/2}}{x^2}} = \lim_{x \to \infty} \frac{3x^{3/2} + \frac{7}{x^{5/2}}}{1 - \frac{1}{x^{3/2}}}$
    • Evaluate limits of individual reciprocal terms as $x \to \infty$:             $\frac{\infty + 0}{1 - 0} = \frac{\infty}{1} = \infty$
  • Example 2 (Algebraic expression involving radical denominators): Evaluate $\lim_{x \to \infty} \frac{4x}{\sqrt{x^2 + 1}}$
    • Direct substitution yields $\left[\frac{\infty}{\infty}\right]$.
    • For positive $x$, $\sqrt{x^2} = x$. Divide numerator and denominator by $x$:             $\lim_{x \to \infty} \frac{\frac{4x}{x}}{\sqrt{\frac{x^2 + 1}{x^2}}} = \lim_{x \to \infty} \frac{4}{\sqrt{1 + \frac{1}{x^2}}} = \frac{4}{\sqrt{1 + 0}} = 4$
    • Boundary Case ($x \to -\infty$): For negative $x$, $\sqrt{x^2} = |x| = -x$. Thus:             $\lim_{x \to -\infty} \frac{4x}{\sqrt{x^2 + 1}} = \lim_{x \to -\infty} \frac{4}{-\sqrt{1 + \frac{1}{x^2}}} = -4$
  • Example 3 (Higher degree denominator): Evaluate $\lim_{x \to \infty} \frac{20x^2 - 3x}{3x^5 - 4x^2 + 5}$
    • Focus on highest powers in denominator ($x^5$). Divide all terms by $x^5$:             $\lim_{x \to \infty} \frac{\frac{20x^2}{x^5} - \frac{3x}{x^5}}{\frac{3x^5}{x^5} - \frac{4x^2}{x^5} + \frac{5}{x^5}} = \lim_{x \to \infty} \frac{\frac{20}{x^3} - \frac{3}{x^4}}{3 - \frac{4}{x^3} + \frac{5}{x^5}}$
    • Evaluate each reciprocal limit as $x \to \infty$:             $\frac{0 - 0}{3 - 0 + 0} = \frac{0}{3} = 0$
    • Comparison with finite limit ($x \to 0$):$\lim_{x \to 0} \frac{20x^2 - 3x}{3x^5 - 4x^2 + 5} = \frac{0 - 0}{0 - 0 + 5} = \frac{0}{5} = 0$