Finding the limit as $x$ approaches negative infinity of $\frac{x}{\sqrt{x^2+1} }$

$\displaylines{\lim_{x\to -\infty} \frac{x}{\sqrt{x^2+1} } }$

The answer my textbook provides is:

$\displaylines{\lim_{x\to -\infty} \frac{x}{\sqrt{x^2+1} } = \lim_{x\to -\infty} \frac{\frac{x}{-x}}{\frac{\sqrt{x^2+1}}{-x} }=\lim_{x\to -\infty} \frac{-1}{\sqrt{1+\frac{1}{x^2} } }= \frac{-1}{\sqrt{1+0} }=-1}$

However, I managed to take a different approach but even though I don't see anything mathematically wrong with it, the answer I'm getting is different:

$\displaylines{\lim_{x\to -\infty} \frac{x}{\sqrt{x^2+1} }=\lim_{x\to -\infty} \frac{x}{\sqrt{x^2(1+\frac{1}{x^2})} }=\lim_{x\to -\infty} \frac{1}{\sqrt{1+\frac{1}{x^2}} }=\frac{1}{\sqrt{1+0} } =1}$.

What am I missing?

Solution 1:

Note that $\sqrt{x^2} = |x|$, so that $\frac{x}{|x|} = \mathrm{sign}(x)$. After that, your calculations should carry through.