Find the limit as x approaches infinity of (e^-x)lnx and (sin2x)/x I'm not sure how to evaluate the limit as $x \rightarrow \infty $ of the following:

$$ \bullet  \lim_{x\to\infty}e^{-x}\cdot \log x $$
$$\bullet\lim_{x\to\infty}  \frac{\sin(2x)}{x}$$ 

 A: Your first expression can be written as
$$\frac{\ln x}{e^x}$$
Both numerator and denominator approach $+\infty$ as $x\to +\infty$, so we can use L'Hôpital's rule.
$$\begin{align}
\lim_{x\to+\infty}\frac{\ln x}{e^x}
 &= \lim_{x\to+\infty}\frac{\frac{d}{dx}\ln x}{\frac{d}{dx}e^x} \\[2 ex]
 &= \lim_{x\to+\infty}\frac{\frac 1x}{e^x} \\[2 ex]
 &= \lim_{x\to+\infty}\frac{1}{xe^x} \\[2 ex]
 &= 0
\end{align}$$
That last equality is due to the numerator being bounded and the denominator approaching $+\infty$.
In your second expression, the absolute value of the numerator $\sin 2x$ is bounded by $1$, while the denominator approaches $+\infty$. This is like the last equality of the previous limit, therefore the limit is zero here also.
A: We can also proceed without using L'Hospital's Rule.  Instead we note the inequalities 
$$\log x<x-1 \tag 1$$ 
and for $x\ge0$
$$e^x\ge \left(1+\frac x2\right)^2 \tag 2$$
both of which can be shown by defining $e^x$ as 
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$$
Then, we have
$$\begin{align}
e^{-x}\log x&=\frac{\log x}{e^x}\\\\
&\le \frac{x-1}{\left(1+\frac x2\right)^2}\\\\
&\to 0\,\,\text{as}\,\,x\to \infty
\end{align}$$

For the second limit, we have 
$$\left|\frac{\sin 2x}{x}\right|\le\frac{1}{x}\to 0\,\,\text{as}\,\,x\to \infty$$
