Understanding $ \frac {e^x}{x^2} $ as $ x \to -\infty $ Given the limit of this function
$$
\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x} = 
\begin{bmatrix}
     \frac {\infty}{\infty} 
\end{bmatrix} = \lim_{x\to -\infty} \frac {e^x (1 + \frac{x^2}{e^x} + \frac{1}{e^x})} {x^2(1 + \frac{1}{x} \sin x)}
$$  
Since $ \lim_{x\to -\infty} e^x = 0$ and $ \lim_{x\to -\infty} x^2 = +\infty$   ,  I'd give the result $ \frac{0}{+\infty} = 0$ but it is not correct.  
What am I mistaking? Have I made any wrong assumption?

Arithmetic of limits:
$$\lim_{x\to -\infty}\frac{e^x}{x^2}=\lim_{x\to-\infty}e^x\cdot\lim_{x\to -\infty}\frac1{x^2}=0\cdot 0= 0$$
And about the original limit:
$$\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x}=\lim_{x\to -\infty}\frac{1+\frac{e^x}{x^2}+\frac1{x^2}}{1+\frac{\sin x}x}=\frac{1+0+0}{1+0}=1$$
Observe that in the limit, $\;x^2\;$ rules, not $\;e^x\;$ .
