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?
 A: 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. 
A: HINT:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\ge \frac{x^2}{2!}$$for $x\ge 0$
A: The biggest mistake in your working is assuming that $0 \cdot \infty = 0$. It is in fact an indeterminate form, so the answer isn't so simple.
One way of computing this would be
$$\lim_{x\to\infty} \frac{x e^x}{x^2+1} = \lim_{x\to\infty} \frac{e^x}{x+\frac{1}{x}}$$
$$\geq \lim_{x\to\infty} \frac{e^x}{x+1}$$
$$=\infty$$
Since $\frac{1}{x}\leq1$, and the exponential function dominates any polynomial.
A: A slight different approach to S.C.B is to note that $e^x>1+x+\frac{x^2}{2}$ and so (worded informally) replacing the e-power in your limit and retaking the limit yields a cubic numerator and a quadratic denominator. Now $x$ goes to infinity, so...
A: Unlike the well known standard limit $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ the following property of exponential function is not that popular/advertized: $$\lim_{x \to \infty}\frac{x^{n}}{e^{x}} = 0, n \in \mathbb{R}$$ It is this particular result which we need here. We have $$\lim_{x \to \infty}\frac{x}{x^{2} + 1}\cdot e^{x} = \lim_{x \to \infty}\frac{x^{2}}{x^{2} + 1}\cdot\frac{e^{x}}{x} = \infty$$
