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Obviously, when substituted, expression gives $\frac{0}{0}$, which leads me to conclusion to use L'Hopital's rule.

However, it is again $\frac{0}{0}$. How can I find this limit?

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  • $\begingroup$ Use L'Hospital 4 times. Tedious, but you will get an answer $\endgroup$
    – Qwerty
    Jul 6, 2016 at 20:53
  • $\begingroup$ Three times should suffice, I believe.. $\endgroup$ Jul 6, 2016 at 20:55
  • $\begingroup$ gave up the second time...thanks! $\endgroup$
    – J.Doe
    Jul 6, 2016 at 20:55
  • $\begingroup$ You must use L'Hospital Rule as many times as needed. But make sure the conditions are met; otherwise you could get the wrong answer. $\endgroup$ Jul 7, 2016 at 20:29

5 Answers 5

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since $\lim _{ x\rightarrow 0 }\frac { e^x -1 }{ x } =1 $ we have

$$\lim _{ x\to 0 } \frac { e^x -e^{ \sin x } }{ x-\sin x } =\lim _{ x\to 0 } \frac { e^{ \sin x } \left( e^{ x-\sin x }-1 \right) }{ x-\sin x } =\lim _{ x\rightarrow 0 }{ e^{ \sin x } } =1 $$

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By the mean value theorem,

$$\frac{e^x - e^{\sin x}}{x-\sin x} = e^{c_x},$$

where $c_x$ is between $x$ and $\sin x.$ As $x\to 0,$ $\sin x \to 0,$ hence $c_x\to 0.$ Thus the limit is $e^0 =1.$

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  • $\begingroup$ This essentially shows the more general result: If $\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=0$ (but $f(x)\ne g(x)$ for $x\ne 0$) and $h'$ exists around $0$ and is continuous at $0$, then $\lim_{x\to 0}\frac{h(f(x))-h(g(x))}{f(x)-g(x)}=h'(0)$. $\endgroup$ Jul 6, 2016 at 21:13
  • $\begingroup$ Beautiful application of MVT! $\endgroup$
    – Qwerty
    Jul 6, 2016 at 21:14
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The expression equals $e^{\sin x} \cdot \frac{e^{x-\sin x}-1}{x-\sin x}.$ The first factor trivially goes to $1$; the second does too, as it can be seen setting $y=x-\sin x$.

Edit: Setting $h=\sin x-x$ immediately yields $$\lim\limits_{x,h\to0}\dfrac{e^{x+h}-e^x}{h},$$ which equals the derivative of $e^x$ at $0$, i.e. $e^0=1$.

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Here is a solution, rewrite $$\lim_{x\to 0} \frac{e^x-e^{\sin x}}{x-\sin x}=\lim_{x\to 0} \frac{e^{x-\sin x}-1}{x-\sin x}e^{\sin x}$$ Notice that as $x\to 0$ we have $e^{\sin x} \to 1$. As $x\to 0$ we also have $x-\sin x\to 0$ also. Therfore our limit is now $$\lim_{u \to 0}\frac{e^{u}-1}{u}$$ Here you can use the taylor expansion for $e^u$ or just L'Hoptial if you wish. Your limit is $1$.

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  • $\begingroup$ The substitution is justified because $f(x)=x-\sin x$ is increasing in a neighborhood of $0$, hence invertible. $\endgroup$
    – egreg
    Jul 7, 2016 at 10:21
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Note that $$\frac{x^m-y^m}{x-y}=\sum_{j=0}^{m-1}x^{m-1-j}y^j$$ and also recall that $$e^y=\sum_{k=0}^{\infty}\frac{y^k}{k!}$$ therefore $$\frac{e^x-e^{\sin x}}{x-\sin x}=1+\sum_{m=2}^{\infty}\frac{\sum_{j=0}^{m-1}x^{m-1-j}\sin^jx}{m!}$$ from which we readily see the limit.

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