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.

$$
\displaylines{
  \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {1 + x^2 }  - x}}{x} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)x^2 }  - x}}{x} \cr 
   = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  - x}}{x} \cr 
   = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  - x}}{x} \cr 
   = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\left( {\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  + 1} \right)}}{x} = -\mathop {\lim }\limits_{x \to  - \infty } \sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  + 1 =- 2 \cr}
$$
