Calculate $\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$ I need to calculate:
$$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$$
I get $0/0$ and can then use l'hopital's rule to find the limit, I can do this but someone asked me how I can do this without using l'hopital's rule. I guess I have to seperate $(x-a)$ in the nominator and denominator. The nominator can be written as $(x-a)(x+a)$ but I don't see how to seperate $(x-a)$.
 A: Hint. Note that for $x\not=a$,
$$\frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}=\frac{(x+2a)(x-a)(\sqrt{2x^2 - ax} +a)}{(2x+a)(x-a)}.$$
A: $$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}\times \frac{\sqrt{2x^2 - ax} +a}{\sqrt{2x^2 - ax} +a}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{2x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{(x-a)(x+a)+a(x-a)}{(x-a)(x+a)+x(x-a)}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\$$simplify $x-a$
$$
\lim_{x \rightarrow a} \frac{(x+a)+a}{(x+a)+x}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\frac{3a}{3a}\times (\sqrt{2a^2 - a^2} +a)=\sqrt{ a^2} +a$$you can apply  l'hopital's rule in this part $$\lim_{x \rightarrow a} \frac{x^2 + ax -2 a^2}{2x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}\cdot \lim_{x \rightarrow a}  \frac{\sqrt{2x^2 - ax} +a}{1}=\\
(\sqrt{ a^2} +a).\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}=\\(|a|+a).\underbrace{\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}}_{l'hopital's rule }=\\(|a|+a).\lim_{x \rightarrow a} \frac{2x + a}{4x-a} =(|a|+a).\frac{3a}{3a}$$
A: You want
$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}
$.
Let
$x = y+a$.
Then
$\begin{array}\\
\dfrac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}
&=\dfrac{(y+a)^2 + a(y+a) - 2a^2}{\sqrt{2(y+a)^2 - a(y+a)} -a}\\
&=\dfrac{y^2+2ay+a^2 + ay+a^2 - 2a^2}{\sqrt{2y^2+4ya+2a^2 - ay-a^2} -a}\\
&=\dfrac{y^2+3ay}{\sqrt{2y^2+3ya+a^2} -a}\\
&=\dfrac{y(y+3a)}{\sqrt{2y^2+3ya+a^2} -a}
\dfrac{\sqrt{2y^2+3ya+a^2} +a}{\sqrt{2y^2+3ya+a^2} +a}\\
&=\dfrac{y(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{2y^2+3ya+a^2 -a^2}\\
&=\dfrac{y(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{2y^2+3ya}\\
&=\dfrac{y(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{y(2y+3a)}\\
&=\dfrac{(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{(2y+3a)}\\
&\to \dfrac{(3a)(\sqrt{a^2} +a)}{(3a)}
\qquad\text{as } y \to 0\\
&= a+|a|
\qquad\text{since } \sqrt{z^2} = |z|\\
\end{array}
$
