<s>[INST] 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.
 [/INST] 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
$$\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}\frac{\sqrt{y^2+a^2}+y}{\sqrt{y^2+a^2}+y}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+b^2}+y}=\frac{a^2}{b^2}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+a^2}+y}\\=\frac{a^2}{b^2}\frac{\sqrt{1+b^2/y^2}+1}{\sqrt{1+a^2/y^2}+1}\to \frac{a^2}{b^2}$$
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