Limit laws, cancellation, and rationalization

Understand the idea

For continuous elementary expressions, substitution is efficient wherever the expression is defined. A $\displaystyle 0/0$ result signals an indeterminate form, not a value. Factor and cancel only on a punctured neighborhood, or multiply by a conjugate. The simplified expression has the same nearby values even when the original has a hole. Check left and right behavior if an absolute value or piecewise rule changes at the approach point.

\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim f(x)}{\lim g(x)}\quad\text{if }\lim g(x)\ne0

Worked example

Evaluate $\displaystyle \lim_{x\to3}(x^2-9)/(x-3)$ .

For

\displaystyle x\ne3

, factor

\displaystyle x^2-9=(x-3)(x+3)

and cancel

\displaystyle x-3

. The limit is

\displaystyle \lim_{x\to3}(x+3)=6

. Cancellation changes the domain at one point but does not change this limit.

Try it yourself

Find $\displaystyle \lim_{x\to0}(\sqrt{4+x}-2)/x$ .

Show solution and reasoning

Rationalizing gives $\displaystyle 1/(\sqrt{4+x}+2)$ for $\displaystyle x\ne0$ , so the limit is $\displaystyle 1/4$ .

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