Evaluating limits with variable exponent and an exponent function with fraction I face a problem when trying to evaluate the following limits. 

$$\lim\limits_{x\rightarrow 0} \left(\frac 14\left(\dfrac{e^2}{x} +3\right)\right)^{4x}.$$
Please help. Thank you.
Assume the limit exists and has value $L$.  Then

$$\begin{align}\log{L} &amp;= \lim_{x \to 0} 4 x \log{\left ( \frac{e^2/x + 3}{4}\right)}\\ &amp;=\lim_{x \to 0} 4 x\log{\frac{e^2}{4 x}} \\ &amp;= \lim_{x \to 0} 4 x (2 - \log{4 x}) \\ &amp;= 0 \end{align}$$

because $\lim_{y \to 0} y \log{y} = 0$.  Then $L=1$.
If $x$ is not an exponent, this is how you do it: let $h=1/x$. The problem now becomes $limit$ $as$ $h$ $approaches$ $infinity$ $of$ $(h*e^2+3)^4/(4^4*h)$. This gives the indeterminate form of $infinity/infinity$, so let's apply L'hopitals rule. $Limit$ $as$ $h$ $approaches$ $infinity$ $of$ $[4e(h*e+3)^3]/4^4$. This should be clear now that the limit approaches $infinity$.