Find $\lim_{x \rightarrow \infty}(\frac{x}{x^2+1}\cdot e^{x})$ I tried:
$\lim_{x \rightarrow \infty}(\frac{x}{x^2+1}\cdot e^{x}) = \frac{1}{x^2+1} \cdot e^xx = 0 \cdot \infty = 0$
But this is wrong. What did I do wrong? How do I solve this?

You're incorrect, as $0 \times \infty \neq 0$. It is indeterminate. 
Note that as $e^t \ge t+1$ for all $t$, we have that $e^{\frac{x}{2}} \ge \frac{x}{2}+1$. Squaring both sides, we have that $$e^{x} \ge \frac{x^2}{4}+x+1>\frac{x^2}{4}+\frac{1}{4} $$
So, note that $$\frac{x}{x^2+1}e^x > \frac{x}{x^2+1} \times \left( \frac{x^2}{4}+\frac{1}{4}\right)=\frac{x}{4}$$
So $$\lim_{x \rightarrow \infty}\left(\frac{x}{x^2+1}\cdot e^{x}\right) \ge \lim_{x \to \infty}\frac{x}{4}$$
However, you can see that the limit on the right tends to infinity. So your limit will also tend to infinity. 
