### Limits with cancellation

In simple cases, one can calculate a limit by just pluggin in the limit value.
But sometimes things ‘blow up’ when the limit number is substituted:
$$\lim_{x\rightarrow 3} {x^2-9\over x-3}={0\over 0}\;\;?????$$
Ick. This is not good. However, in this example, as in *many*
examples, doing a bit of simplifying algebra first gets rid of the
factors in the numerator and denominator which cause them to vanish:
$$\lim_{x\rightarrow 3} {x^2-9\over x-3}=
\lim_{x\rightarrow 3} {(x-3)(x+3)\over x-3}=
\lim_{x\rightarrow 3} {(x+3)\over 1}={(3+3)\over 1}=6$$
Here at the very end we *did* just plug in, after all.

The lesson here is that some of those darn algebra tricks
(‘identities’) are helpful, after all. If you have a ‘bad’ limit, *always* look for some *cancellation* of factors in the numerator
and denominator.

In fact, for hundreds of years people *only* evaluated limits
in this style! After all, human beings can't really execute infinite
limiting processes, and so on.

#### Exercises

- Find $\lim _{x\rightarrow 2} { x-2 \over x^2-4 }$
- Find $\lim _{x\rightarrow 3} { x^2-9 \over x-3 }$
- Find $\lim _{x\rightarrow 3} { x^2 \over x-3 }$

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##### Calculus Refresher

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