Precalculus / Unit 10: Sequences, Series and Intro to Limits
10.5 Limits Intuitively
Notes
The limit $\displaystyle \lim_{x \to a} f(x) = L$ means $f(x)$ gets close to $L$ as $x$ gets close to $a$ (from both sides), whether or not $f(a)$ exists.
- For polynomials, substitute: $\displaystyle \lim_{x \to 2}(3x^2 - x) = 12 - 2 = 10$.
- If substitution gives $\displaystyle \frac{0}{0}$, simplify first (factor or rationalize), then substitute.
- Example: $\displaystyle \lim_{x \to 3}\frac{x^2 - 9}{x - 3} = \lim_{x \to 3}(x + 3) = 6$.
Practice
Find $\displaystyle \lim_{x \to 2}(3x^2 - x)$.
Substitute $x = 2$: $12 - 2 = 10$.
Find $\displaystyle \lim_{x \to 3}\frac{x^2 - 9}{x - 3}$.
Factor and cancel: $\displaystyle \frac{(x - 3)(x + 3)}{x - 3} = x + 3$, which approaches 6.
Find $\displaystyle \lim_{x \to 4}\frac{\sqrt{x} - 2}{x - 4}$.
Since $\displaystyle x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$, the expression equals $\displaystyle \frac{1}{\sqrt{x} + 2}$, which approaches $\displaystyle \frac{1}{4}$.
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