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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.

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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 $x = a - \epsilon$.

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  • $\begingroup$ then we get a+x in numerator and (-1)*(x-a) in denominator, which results still in zero. $\endgroup$ – drin Mar 22 '15 at 21:10
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    $\begingroup$ so this means there is no limit then, thanks alot. $\endgroup$ – drin Mar 22 '15 at 21:21
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Assume that $a\neq 0 \Rightarrow \dfrac{a^2-x^2}{(x-a)^2} = -\dfrac{x+a}{x-a} \to \text{ limit does not exist}$

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  • $\begingroup$ How did you get x+a in numerator? $\endgroup$ – drin Mar 22 '15 at 21:00
  • $\begingroup$ you divide both the top and bottom by $x-a$ $\endgroup$ – DeepSea Mar 22 '15 at 21:00
  • $\begingroup$ But how do you get (x-a)*(x+a) in numerator, when we have a^2-x^2, and as far as i know a-x is not the same as x-a. $\endgroup$ – drin Mar 22 '15 at 21:02
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    $\begingroup$ Difference of squares: $a^2 - b^2 = (a - b)(a + b)$ $\endgroup$ – Axoren Mar 22 '15 at 21:41
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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.

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