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.

Hint: Multiply numerator and denominator by $\sqrt{1+x^2} + x$.
$$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)\cdot \frac{\sqrt{1+x^2} +x}{\sqrt{1 + x^2}+x} = \lim_{x\to -\infty}\frac{1 + x^2 - x^2}{x(\sqrt{1 + x^2} + x)}$$
$$= \lim_{x\to -\infty}\frac 1{x\sqrt{1 + x^2} + x^2}$$
Can take it from here?
