1.5 Algebraic Techniques
Notes
Direct substitution that gives $\displaystyle \frac{0}{0}$ is an indeterminate form. It means more work is needed.
- Factor and cancel: $\displaystyle \frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$, so the limit at 3 is 6.
- Multiply by the conjugate when square roots are involved.
- If substitution gives a nonzero number over 0, such as $\displaystyle \frac{5}{0}$, the limit does not exist as a finite number (the function grows without bound).
- Example: $\displaystyle \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} = \frac{1}{4}$.
Practice
Find $\displaystyle \lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.
Factor: $\displaystyle \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \to 6$.
Find $\displaystyle \lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}$.
$x^2 + 5x + 6 = (x + 2)(x + 3)$, so the expression becomes $x + 3 \to 1$.
Find $\displaystyle \lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$.
Multiply top and bottom by $\displaystyle \sqrt{x + 9} + 3$ to get $\displaystyle \frac{x}{x(\sqrt{x + 9} + 3)} \to \frac{1}{6}$.
Ready for more? Take the Unit 1 test.