Find the limit of fuction. Find the Limit: $$ \lim_{x \to a} \frac{a^2-x^2}{(x-a)^2}$$
I tried solving this problem, and while on that, I got in numerator $(a-x)(a+x)$ and in denominator i had $(x-a)(x-a)$, but I just can't go on any further, I guess all I need is to get rid of all the denominator since it gives me zero which I should avoid, but I can't find a way to cancel them out. I would be very thankful for your help.
 A: Assume that $a\neq 0 \Rightarrow \dfrac{a^2-x^2}{(x-a)^2} = -\dfrac{x+a}{x-a} \to \text{ limit does not exist}$
A: Assume $a\neq0$, we have, for $x \neq a$,
$$
\frac{a^2-x^2}{(a-x)^2}=\frac{(a-x)(a+x)}{(a-x)(a-x)}=\frac{a+x}{a-x}
$$ and 
$$
\frac{a+x}{a-x} \to
\begin{cases}
\text{sign($a$)} \times \infty,  & \text{if $x \to a^-$} \\
-\text{sign($a$)} \times \infty, & \text{if $x \to a^+$.}
\end{cases}
$$ There is no limit.
A: Multiply the denominator by $(-1)(-1)=1$. Take one minus sign to turn $(x-a)$ into $(-x+a)$.
Or do a similar trick in the numerator.
For example:
$$\frac{(a^2 - x^2)}{(x-a)^2} = \frac {(a + x)(a-x)}{(x-a)(x-a)}$$
$$ =  \frac {(a + x)(a-x)}{(-1)(-x+a)(x-a)} = -\frac {(a+x)(a-x)}{(a-x)(x-a)}$$
$$ = -\frac {a+x}{x-a} = \frac {a+x}{a-x}$$
Edit: just to see why there's no limit,
Let $\epsilon > 0$ be some positive number. Suppose $x = a + \epsilon$, meaning, $x$ is a little bigger than $a$, then:
$$\frac {a+(a + \epsilon)}{a-(a + \epsilon)} = \frac {2a + \epsilon}{-\epsilon} = -\frac {2a}{\epsilon} - 1$$
Now make epsilon really small to see what happens to the function "just to the right" of $x = a$. Meaning, you are analyzing the limit as $x \to a^+$.
A similar analysis will let you see what happens when $x$ is is just a little smaller than $a$, to analyze $x \to a^-$, by putting $x = a - \epsilon$.
