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I have problem evaluating limits with the variable in power, like the following limits:

  • $\lim_{x \to 0} (1+ \sin 2x)^{\frac{1}{x}}$
  • $\lim_{x \to \infty} \big(\frac{2x+5}{2x-1})^{2x}$

I asked the question like this to get the main idea behind evaluating these kind of limits, so I can solve all the related questions.

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Wolfram alpha does an excellent job of explaining this. wolframalpha.com/input/… and ask to show steps. –  Galois Group Jan 25 '12 at 8:18
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math.stackexchange.com/questions/47230/… The question in the link explains how to solve this kind of limits. –  Beni Bogosel Jan 25 '12 at 8:21
    
I guess we had quite a few question (and answers) of similar type, e.g. these two: $\lim\limits_{x\to\infty}\left(\frac{x}{x-1}\right)^{2x+1}$ here, $\lim \limits_{x\to \infty}(e^{2x}+1)^{1/x}$ here. And of course, the generalization from the post linked in Beni's comment gives a very good explanation what to do in general. –  Martin Sleziak Jan 25 '12 at 9:29

2 Answers 2

up vote 4 down vote accepted

A good idea whenever the exponent is variable is to take logarithims. So for example, to compute the first limit first find the limit of $$ \log \left( 1+\sin 2x \right)^{1/x} = \frac{1}{x} \cdot \log (1+ \sin 2x) .$$ As $x\to 0$ this is of the form $0/0$ so by applying L'Hopitals rule we find that the limit is equal to $ \lim_{x\to 0} \frac{ 2\cos 2x}{1+\sin 2x} = 2$, which in turn implies the original limit is $e^2.$

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For evaluating the first limit (this requires a Taylor series), use

$$(1+\sin2x)^{1/x}=\exp\left(\frac{\log(1+\sin2x)}{x}\right); \tag{A}$$

$$\log(1+u)=u-\frac{1}{2}u^2+\frac{1}{3}u^3-\cdots; \tag{B}$$

$$\lim_{x\to0}\frac{\sin x}{x}=1. \tag{C}$$

As for the second, write $u=2x-1$ and the expression becomes

$$\lim_{u\to\infty}\left(1+\frac{4}{u}\right)^{u+1}. \tag{D}$$

You should hopefully be able to resolve that one on your own.

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