Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be something easy I just can't see.
Thanks in advance.
 A: Let $y=-x\to \infty$ then
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}=\lim_{y\rightarrow  \infty}\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}$$
and
$$\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}\frac{\sqrt{y^2+a^2}+y}{\sqrt{y^2+a^2}+y}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+b^2}+y}=\frac{a^2}{b^2}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+a^2}+y}\\=\frac{a^2}{b^2}\frac{\sqrt{1+b^2/y^2}+1}{\sqrt{1+a^2/y^2}+1}\to \frac{a^2}{b^2}$$
A: Hint:
Do divide top and bottom by $x$ as you have tried, using that for negative values of $x$ (remember that $x\to -\infty$): $$x=-\sqrt{x^2}$$
A: We can rewrite the fraction as follows
\begin{align*}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}\times\frac{\sqrt{x^2+a^2}-x}{\sqrt{x^2+a^2}-x}\times \frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+b^2}-x}\\[4pt]
&=\frac{x^2+a^2-x^2}{x^2+b^2-x^2}\times\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x}
\end{align*}
So
\begin{align*}
\lim_{x\to-\infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}&=\frac{a^2}{b^2}\times\lim_{x\to-\infty}\frac{\sqrt{x^2+b^2}-x}{\sqrt{x^2+a^2}-x}\\[4pt]
&=\frac{a^2}{b^2}\times\lim_{x\to-\infty}\frac{\sqrt{\frac{x^2}{x^2}+\frac{b^2}{x^2}}-\frac x{|x|}}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2}}-\frac x{|x|}}\\[4pt]
&=\frac{a^2}{b^2}\times\frac{\sqrt{1}-(-1)}{\sqrt{1}-(-1)}\\[4pt]
&=\frac{a^2}{b^2}
\end{align*}
A: From 
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
one can factor an $x$ from each term as follows:
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{x \left(1 + \sqrt{1 + \frac{a^{2}}{x^2}} \right)}{x \left( 1 + \sqrt{1 + \frac{b^2}{x^2}} \right)} =
 \frac{1 + \sqrt{1 + \frac{a^{2}}{x^2}} }{ 1 + \sqrt{1 + \frac{b^2}{x^2}} }.
\end{align}
From here the limit can be taken or one can expand the one more time. Using 
\begin{align}
\sqrt{1 + t} = 1 + \frac{t}{2} - \frac{t^2}{8} + \mathcal{O}(t^3)
\end{align}
then 
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{2 + \frac{a^2}{2 x^2} - \frac{a^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)}{2 + \frac{b^2}{2 x^2} - \frac{b^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)} = 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right).
\end{align}
Upon taking the limit the result becomes
\begin{align}
\lim_{x \to \pm \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} = \lim_{x \to \pm \infty} 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right) = 1.
\end{align}
