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.
 A: 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 $-x=t\gt 0$.
$$\lim_{x\to -\infty}\frac{\sqrt{1+x^2}-x}{x}=\lim_{t\to\infty}\frac{\sqrt{1+(-t)^2}+t}{-t}=\lim_{t\to\infty}\left(-\sqrt{\frac{1+t^2}{t^2}}-1\right)=-2.$$
A: 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?
A: Note that $\tan t$ maps $(-\pi/2,\pi/2)$ to $(-\infty,\infty)$ and is increasing on this interval. Consequently we can validly substitute $x=\tan t$ and get $$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)=\lim_{t \to -\frac{\pi}{2}^+} \left(\frac{\sqrt{1+\tan^2(t)}-\tan t}{\tan t} \right)$$ which is readily computed if we recall some trigonometry. (Note that we must ensure the correct sign of the square root.)
A: Setting $h=-\dfrac1x,$
$$\lim_{x \to -\infty} \left(\frac{\sqrt{1+x^2}-x}{x} \right)$$
$$=\lim_{h \to 0^+}\left(\frac{\sqrt{1+\dfrac1{h^2}}+\dfrac1h}{-\dfrac1h} \right)$$ 
$$=-\lim_{h \to 0^+}\left(\dfrac{\sqrt{h^2+1}}{|h|}+\dfrac1h \right)h$$
$$=-\lim_{h \to 0^+}\left(\sqrt{h^2+1}+1 \right)\text{ as }h\ne0\text{ as }h\to0^+$$
$$=-\left(\sqrt{0^2+1}+1 \right)$$
A: $$
\displaylines{
  \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {1 + x^2 }  - x}}{x} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)x^2 }  - x}}{x} \cr 
   = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  - x}}{x} \cr 
   = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  - x}}{x} \cr 
   = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\left( {\sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  + 1} \right)}}{x} = -\mathop {\lim }\limits_{x \to  - \infty } \sqrt {\left( {1 + \frac{1}{{x^2 }}} \right)}  + 1 =- 2 \cr}
$$
