Find $\lim \limits_{x\to 8} {\frac{64-x^2}{8-x}}$ I'm making a bunch of limits exercises and I found me stuck with this limit:

$$\begin{align}
\lim_{x\to 8} {\frac{64-x^2}{8-x}} 
\end{align}$$

It gives me an indetermination of 

$$\begin{align} \frac{0}{0} \end{align}$$

So, the next step is to factorize. I used the Ruffini's Rule.

The solution is $16$. And my result was $-16$, something is wrong :(

Any help?
To use Ruffini's Rule, you need to use x-8, not 8-x. So, multiply both numerator and denominators by -1 one and then apply Ruffini's rule. You will get the correct answer, which is +16.
Another trick for futur reference, is to compute the derivatives of both numerators and denominators and then compute the limit. In this case, you will get -2x divided by -1, which gives you 2x. When you replace x with 8, you get the answer +16.

Note: I saw that you said Ruffini's rule gives you -x-8, I think that is where your error is: when you multiply by -1, you should get -(8-x) = x-8. Same applies to the numerator.
$$
\frac{64 - x^2}{8-x} = 8 + x \;,
$$
so, the limit as $x \rightarrow 8$ is of course $16$.