Calculus Limits and Algebraic Techniques Study Notes

Fundamental Concept: Direct substitution into rational functions can lead to indeterminate forms such as $\frac{0}{0}$ . When direct substitution yields an indeterminate form, algebraic factoring allows common terms in the numerator and denominator to be canceled before re-evaluating the limit.
Direct Substitution Example:
- Problem: Evaluate the limit as $x \to 3$ for the expression: $\lim_{x \to 3} \frac{1}{x+3}$
- Procedure:
- Substitute $x = 3$ directly into the denominator: $\lim_{x \to 3} \frac{1}{x+3} = \frac{1}{3+3}$
- Simplify denominator sum: $\frac{1}{3+3} = \frac{1}{6}$
- Final Answer: $\frac{1}{6}$
Factoring Common Zero Factors Example:
- Problem: Evaluate the limit as $x \to 3$ for the expression: $\lim_{x \to 3} \frac{(x-3)(x-1)}{x-3}$
- Procedure:
- Direct substitution of $x = 3$ yields the indeterminate form $\frac{0}{0}$ due to the factor $(x-3)$ in both numerator and denominator.
- Cancel the common factor $(x-3)$ from the numerator and denominator: $\frac{(x-3)(x-1)}{x-3} = \frac{x-1}{1}$
- Rewrite the limit statement with the reduced expression: $\lim_{x \to 3} \frac{x-1}{1}$
- Substitute $x = 3$ into the simplified expression: $\frac{3-1}{1} = 2$
- Final Answer: $2$

Fundamental Concept: When determining the limit of a rational function as $x \to \infty$ , divide every term in both the numerator and denominator by the highest power of $x$ present in the denominator.
Limit Behavior at Infinity Rule:
- For any positive power $p > 0$ and constant $c$ : $\lim_{x \to \infty} \frac{c}{x^p} = 0$
Step-by-Step Problem Solution:
- Problem: Evaluate the limit at infinity: $\lim_{x \to \infty} \frac{\frac{5x^2 - 10}{x^2}}{\frac{2x^2 - x - 12}{x^2}}$
- Procedure:
- Divide each term individually by $x^2$ : $\lim_{x \to \infty} \frac{5 - \frac{10}{x^2}}{2 - \frac{1}{x} - \frac{12}{x^2}}$
- Apply the limit behavior as $x \to \infty$ to each term containing $x$ in the denominator:
  - $\frac{10}{x^2} \rightarrow 0$
  - $\frac{1}{x} \rightarrow 0$
  - $\frac{12}{x^2} \rightarrow 0$
- Substitute these zero limits into the expression: $\frac{5 - 0}{2 - 0 - 0}$
- Simplify the resulting numerical fraction: $\frac{5}{2}$
- Final Answer: $\frac{5}{2}$

Fundamental Concept: When a limit containing a radical produces an indeterminate form $\frac{0}{0}$ , multiply by the radical conjugate to rationalize the expression and eliminate zero-producing factors.
Difference of Squares Identity:
- Rationalization relies on the algebraic identity: $(a-b)(a+b) = a^2 - b^2$
- Applied to radical expression $(\sqrt{4x+1} - 1)$ : $(\sqrt{4x+1} - 1)(\sqrt{4x+1} + 1) = (\sqrt{4x+1})^2 - (1)^2 = 4x + 1 - 1 = 4x$
Step-by-Step Problem Solution:
- Problem: Evaluate the limit as $x \to 0$ : $\lim_{x \to 0} \frac{4x}{x(\sqrt{4x+1} + 1)}$
- Procedure:
- Identify the common factor $x$ present in both the numerator and denominator.
- Cancel factor $x$ from numerator and denominator: $\frac{4x}{x(\sqrt{4x+1} + 1)} = \frac{4}{\sqrt{4x+1} + 1}$
- Substitute $x = 0$ directly into the simplified expression: $\frac{4}{\sqrt{4(0)+1} + 1}$
- Simplify the arithmetic inside and outside the square root: $\frac{4}{\sqrt{0+1} + 1} = \frac{4}{\sqrt{1} + 1} = \frac{4}{1 + 1} = \frac{4}{2} = 2$
- Final Answer: $2$

Fundamental Trigonometric Limit Theorem:
- For any real variable $u$ approaching zero: $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$
- Specifically, for argument $u = 2\theta$ : $\lim_{\theta \to 0} \frac{\sin(2\theta)}{2\theta} = 1$
Step-by-Step Problem Solution:
- Problem: Evaluate the trigonometric limit: $\lim_{\theta \to 0} \frac{2 \sin(2\theta)}{3 \cdot 2\theta}$
- Procedure:
- Separate the constant multiplier $\frac{2}{3}$ from the special trigonometric limit fraction $\frac{\sin(2\theta)}{2\theta}$ : $\frac{2}{3} \cdot \left(\frac{\sin(2\theta)}{2\theta}\right)$
- Apply the limit property $\lim_{\theta \to 0} \frac{\sin(2\theta)}{2\theta} = 1$ : $\frac{2}{3} \cdot (1)$
- Perform the multiplication: $\frac{2}{3}$
- Final Answer: $\frac{2}{3}$