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How do you calculate $\displaystyle \lim_{{x\to 0}}(1+\sin{x})^{\frac{1}{x}}$? I got it from here. It says L'Hopital, but I can't figure out how to apply it as I don't have a denominator. I also tried to rewrite the limit using trig identities:

$\displaystyle \lim_{{x\to 0}}(1+\sin{x})^{\frac{1}{x}} = \lim_{x \to 0} 2^{\frac{1}{x}}\sin^{\frac{2}{x}}\left(\frac{\pi}{4}+\frac{x}{2}\right) = ?$

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    $\begingroup$ Try using the exponential function. The continuity of the exponential allows you to move the limit into its argument. edit: the answer below is more elegant, but my suggestion gives a clear denominator $\endgroup$ – Eric Thoma Jan 3 '16 at 5:00
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Hint: $\dfrac{1}{x} = \dfrac{1}{\sin x}\cdot \dfrac{\sin x}{x}$

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If we wish to use L'Hospital's Rule, then we can write

$$(1+\sin x)^{1/x}=e^{\frac{\log(1+\sin x)}{x}}$$

Then,

$$\begin{align} \lim_{x\to 0}(1+\sin x)^{1/x}&=\lim_{x\to 0}e^{\left(\frac{\log(1+\sin x)}{x}\right)}\\\\ &e^{\lim_{x\to 0}\left(\frac{\log(1+\sin x)}{x}\right)}\\\\ &=e^{\lim_{x\to 0}\left(\frac{\cos x}{1+\sin x}\right)}\\\\ &=e \end{align}$$

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  • $\begingroup$ replace e with 10 or log with ln? :P $\endgroup$ – BCLC Jan 3 '16 at 5:31
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    $\begingroup$ @BCLC What are you asking? Typically, we use $\log (x)$ to denote the natural logarithm. That is a very common convention. We use $\log_{b}(x)$ to denote the base $b\ne e$ logarithm. $\endgroup$ – Mark Viola Jan 3 '16 at 5:33
  • $\begingroup$ lots of books do that :P. idk i think some would find it confusing. i mean why else do we have $\ln(x)$? less characters to type + less confusing $\endgroup$ – BCLC Jan 3 '16 at 5:34
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    $\begingroup$ @BCLC It should be quite clear when one writes $z=e^{\log(z)}$, don't you think? $\endgroup$ – Mark Viola Jan 3 '16 at 5:36
  • $\begingroup$ @Dr.MV. May I confess that, at my age, I only know one kind of logarithms. You could be interested by the fact that when we were teached about decimal logarithms, the official terminology was (in French) common or even vulgar. $\endgroup$ – Claude Leibovici Jan 3 '16 at 8:59
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my answer:

$\lim_{x\to 0}\left(1+\sin x\right)^{1/x}$

$=\lim_{x\to 0}(\left(1+\sin x\right)^{\frac{1}{\sin x}})^{\frac{\sin x}{x}}$

Note: $\lim_{x\to 0}\frac{\sin x}{x}=1$, so i get

$=(e)^1=e$

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