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I need to solve the following limit without using L'Hopital's rule:

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

The thing is that I can not figure out what to do. One of my ideas was to apply this rule: $a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$, getting this:

$$\lim _{x\to 0}e^{\frac{1}{x}\ln \left(1+\sin \left(x\right)\right)}$$

I already know that the answer is $e$, so the exponent is definitely 1. However, I tried everything I could but have no idea how to solve $\lim _{x\to 0}\left(\frac{1}{x}\ln \left(1+\sin \left(x\right)\right)\right)$ which needs to be 1.

I would really appreciate your help, and if you find a totally different way to solve the limit without using L'Hospital's rule it will be good as well.

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We have

$$\left(1+\sin x\right)^{\frac{1}{x}}=\left[\left(1+\sin x\right)^{\frac{1}{\sin x}}\right]^{\frac{\sin x}{x}}\to e^1=e$$

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    $\begingroup$ A usual comment: the limit holds (of course), but you need you justify it. Intuitive things like $a_n^{b_n}\to a^b$ are not obvious... $\endgroup$ – Clement C. Jul 11 '18 at 15:43
  • $\begingroup$ @ClementC. Yes of course, by $$f(x)^{g(x)}=e^{g(x)\log f(x)}$$ we can see that the limit holds. $\endgroup$ – gimusi Jul 15 '18 at 8:44
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Consider the function $f\colon(-\frac{\pi}{2},\frac{\pi}{2})\to(-1,1)$ defined by $$ f(x) = \ln(1+\sin x)\,. $$

We have that $f(0)=0$, $f$ is continuously differentiable, and $f'(x) = \frac{\cos x}{1+\sin x}$ by the chain rule; so that $$ \lim_{x\to 0} \frac{\ln(1+\sin x)}{x}=\lim_{x\to 0} \frac{f(x)-f(0)}{x} = f'(0) = \frac{\cos 0}{1+\sin 0} = 1 $$ so that by continuity of $\exp$ $$ \lim_{x\to 0} e^{\frac{1}{x}\ln(1+\sin x)} = e^{\lim_{x\to 0} \frac{1}{x}\ln(1+\sin x)} = e^1 = e\,. $$

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Expand the term in the exponent with the help of Taylor series for $\ln(1+x)$ where $x\lt\lt1$ which is true for this case. $\ln(1+x) = x-x^2/2+x^3/3-\dots $ We see $\lim_{x\to0} \frac{\ln(1+x)}x =1$

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Consider $$A=\left(1+\sin\left(x\right)\right)^{\frac{1}{x}}\implies\log(A)={\frac{1}{x}}\log\left(1+\sin\left(x\right)\right)$$ Now, using Taylor series and compositions of them $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\log\left(1+\sin\left(x\right)\right)=x-\frac{x^2}{2}+O\left(x^3\right)$$ $$\log(A)=1-\frac{x}{2}+O\left(x^2\right)$$ $$A=e^{\log(A)}=e-\frac{e x}{2}+O\left(x^2\right)$$ which shows the limit and how it is approached.

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Hint: Remember, log and exp are continuous functions. So you can take limits in and out. i.e $$\lim_{x \to c} f(x)^{g(x)} = \lim \Big (\exp \big( \log f(x)^{g(x)}\big)\Big) = \lim \Big (\exp \big(g(x) \log f(x)\big)\Big) \\ = \exp \Big( \lim (g(x)) \cdot \log \big(\lim (f(x)) \big) \Big)$$

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

Note: exp denotes here the exponential function. That is, $\exp (x) = e^x$. log is the natural logarithm.

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  • $\begingroup$ You mean we can transfer limit $x\rightarrow 0$ to the power ignoring $e$ ? $\endgroup$ – Entrepreneur Jul 7 '18 at 6:03
  • $\begingroup$ @Entrepreneur Yes, Essentially. $\endgroup$ – Praneet Srivastava Jul 7 '18 at 6:20

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