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 $(x-a)$.

$$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}\times \frac{\sqrt{2x^2 - ax} +a}{\sqrt{2x^2 - ax} +a}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{2x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{(x-a)(x+a)+a(x-a)}{(x-a)(x+a)+x(x-a)}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\$$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 $$\lim_{x \rightarrow a} \frac{x^2 + ax -2 a^2}{2x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}\cdot \lim_{x \rightarrow a}  \frac{\sqrt{2x^2 - ax} +a}{1}=\\
(\sqrt{ a^2} +a).\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}=\\(|a|+a).\underbrace{\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}}_{l'hopital's rule }=\\(|a|+a).\lim_{x \rightarrow a} \frac{2x + a}{4x-a} =(|a|+a).\frac{3a}{3a}$$
