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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.

Srivatsan
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Gigili
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  • Wolfram alpha does an excellent job of explaining this. http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Cto+0%7D+%281%2B+%5Csin+2x%29%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D and ask to show steps. – Arkady Jan 25 '12 at 08:18
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    http://math.stackexchange.com/questions/47230/value-of-lim-n-to-infty1-frac2n2-cosnn3nn/47238#47238 The question in the link explains how to solve this kind of limits. – Beni Bogosel Jan 25 '12 at 08:21
  • 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 09:29

2 Answers2

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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 $e^2.$

Ragib Zaman
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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.

anon
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