# Math Insight

### 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

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