limits - Evaluate $lim_{x to -infty} left(frac{sqrt{1+x^2}-x}{x} right)$

Evaluate $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$

I tried by taking $x^2$ out of the root by taking it common.

i.e: $$\lim_{x \to -\infty} \left(\frac{x\sqrt{\frac{1}{x^2}+1}-x}{x} \right)$$
and then cancelling the x in numerator and denominator

$$\lim_{x \to -\infty} \left(\frac{\sqrt{\frac{1}{x^2}+1}-1}{1} \right)$$

then substituting $x= -\infty$ in the equation, we get,
$$\lim_{x \to -\infty} \left(\frac{\sqrt{0+1}-1}{1} \right)$$
which equals to $0$. But it is not the correct answer.

What have I done wrong.

Answer

When $x\lt 0$, we have
$$\sqrt{1+x^2}\not =x\sqrt{\frac{1}{x^2}+1}.$$
(Note that LHS is positive and that RHS is negative!)

You can set $-x=t\gt 0$.
$$\lim_{x\to -\infty}\frac{\sqrt{1+x^2}-x}{x}=\lim_{t\to\infty}\frac{\sqrt{1+(-t)^2}+t}{-t}=\lim_{t\to\infty}\left(-\sqrt{\frac{1+t^2}{t^2}}-1\right)=-2.$$