Asymptotes of the graph of $\sqrt{1+x^2}+2x$ against $x$

Question

I am computing the asymptotes of this function: $$\sqrt{1+x^2}+2x$$

When I try to to the oblique asymptote ($y=mx+n)$ and by doing the limits I got only one solution: $y=3x$. But the solution also gives $y=x$ as an answer. I don't know how $$\lim_{x\to\infty} \frac{\sqrt{1+x^2}+2x}{x} = 1$$

I tried to compute it multiple times but I only get $3$ as a result.

Any help?

Answer 1

We have $$\lim_{x\,\to\,+\infty}\frac{\sqrt{1+x^2}+2x}x\ =\ \lim_{x\,\to\,+\infty}\sqrt{\frac{1+x^2}{x^2}}+2\ =\ 3$$ but $$\lim_{x\,\to\,-\infty}\frac{\sqrt{1+x^2}+2x}x\ =\ \lim_{x\,\to\,+\infty}\frac{\sqrt{1+(-x)^2}+2(-x)}{-x}\ =\ \lim_{x\,\to\,+\infty}-\sqrt{\frac{1+x^2}{x^2}}+2\ =\ 1$$

Answer 2

if you meant $x$ tends to $+\infty$ we can write $$\frac{x\left(\sqrt{1+\frac{1}{x^2}}+2\right)}{x}$$

Answer 3

$$\lim\limits_{x\to \pm\infty} \frac{\sqrt{1+x^2}+2x}{x}= \lim\limits_{x\to \pm\infty} \frac{|x|}{x}\sqrt{\frac{1}{x^2}+1}+2 = \lim\limits_{x\to \pm\infty} \pm\sqrt{\frac{1}{x^2}+1}+2 =\lim\limits_{x\to \pm\infty} \pm1+2$$

Thus the limit is $3$ for $x\to +\infty$ and $1$ for $x\to-\infty$