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?

Why not:
$$
\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x} = 
\begin{bmatrix}
     \frac {\infty}{\infty} 
\end{bmatrix} = \lim_{x\to -\infty} \frac {x^2 (1 + \frac{e^x}{x^2} + \frac{1}{x^2})} {x^2(1 + \frac{1}{x} \sin x)}=1?
$$
