How do I solve this infinite limit? I have this limit:
$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} $$ I tried to solve it by this:
$$ \lim_{x\to\infty}\frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{\frac{x^3}{x^3}+\frac{\cos x}{x^3}+\frac{e^{-2x}}{x^3}}{\frac{x^2\sqrt{x^2+1}}{x^3}} = \frac{0+0+0}{\frac{\sqrt{\infty^2+1}}{\infty}}$$ I do not think that I got it right there... Wolfram also says that the answer is $1$, which this does not seems to be. How do I solve this?

For sufficiently large $x$
$\displaystyle \frac{x^3-1}{x^2\sqrt{x^2+1}} \leq \frac{x^3+\cos x+e^{-2x}}{x^2\sqrt{x^2+1}} \leq \frac{x^3+2}{x^2\sqrt{x^2+1}}$
Now, 
$\displaystyle \frac{2}{x^2\sqrt{x^2+1}} \rightarrow 0$, and  $\displaystyle \frac{-1}{x^2\sqrt{x^2+1}} \rightarrow 0$
Moreover,
$\displaystyle \frac{x^3}{x^2\sqrt{x^2+1}}=\frac{x^3}{x^2\sqrt{x^2(1+\frac{1}{x^2})}}=\frac{x^3}{x^3\sqrt{1+\frac{1}{x^2}}}=\frac{1}{1\sqrt{1+\frac{1}{x^2}}} \rightarrow 1$
