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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.

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The problem as you state it is $$\lim_{x \rightarrow 0} \left ( \frac{\frac{e^2}{x} + 3}{4}\right )^4 x$$ but I think you mean $$\lim_{x \rightarrow 0} \left ( \frac{e^{2/x}+3}{4} \right )^{4 x}$$ – Ron Gordon Apr 15 '13 at 11:57
What problem do you have? How far do you get? Does e^2/x mean $e^{2/x}$ or $(e^2)/x$? – Gerry Myerson Apr 15 '13 at 11:58
What methods have you learned for calculating limits? – Gerry Myerson Apr 15 '13 at 11:58
Anyways, as you have $4x$ as an exponent, I would suggest you use the exponential form: $\displaystyle f(x)=\left(\frac{e^{2/x}+3}{4}\right)^{4x}=e^{4x\cdot\ln\left(\frac{e^{2/x}+3}{4‌​}\right)}=e^{u(x)}$ Then find the limit of $u(x)$ which will give you the limit of $f(x)$ – Dolma Apr 15 '13 at 16:11

Assume the limit exists and has value $L$. Then

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

because $\lim_{y \to 0} y \log{y} = 0$. Then $L=1$.

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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 $infinity$.

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I'm not sure this would be very useful to him since it seems like it does have $x$ as an exponent ... – Dolma Apr 15 '13 at 16:06

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