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Finding this limit using L'Hôpital's rule is easy, but how to do it without using L'Hôpital's rule?

$$\lim_{x \rightarrow 0} \frac{(1+\sin x)^{\csc x}-e^{x+1}}{\sin (3x)}$$

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  • $\begingroup$ Please see the editing. I think there is a mistake here. Did you mean $e^x$ or $e^{x+1}$? $\endgroup$ – Tunk-Fey Jun 23 '14 at 10:25
  • $\begingroup$ @Tunk-Fey, $(1+\sin(x))^{\csc(x)}\to e$ as $x\to 0$. If it were $e^x$, l'Hôpital's rule would not apply. $\endgroup$ – Joffysloffy Jun 23 '14 at 10:26
  • $\begingroup$ @Joffysloffy Sorry, I thought it was equal to $1$. $\endgroup$ – Tunk-Fey Jun 23 '14 at 10:29
  • $\begingroup$ @Tunk-Fey, That's okay :). Note that it is equivalent to the limit definition of $e$ by using substitution of limits. $\endgroup$ – Joffysloffy Jun 23 '14 at 10:30
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We may proceed as follows and reduce the complicated limit expression to a simple one before applying series expansions $$\begin{aligned}L&=\lim_{x \to 0}\frac{(1 + \sin x)^{\csc x} - e^{x + 1}}{\sin 3x}\\ &=\lim_{x\to 0}\frac{\exp\{\csc x\log(1+\sin x)\} -\exp(1+x)}{\sin 3x}\\ &=\lim_{x\to 0}\frac{\exp(1+x)\left\{\exp\left(\csc x\log(1+\sin x) -1 -x\right) -1\right\}}{\sin 3x}\\ &=\lim_{x\to 0}\frac{\exp(1+x)\left\{\exp\left(\csc x\log(1+\sin x) -1 -x\right) -1\right\}}{3x}\cdot\frac{3x}{\sin 3x}\\ &=\frac{e}{3}\lim_{x\to 0}\frac{\exp\{\csc x\log(1+\sin x) -1 -x\} -1}{x}\\ &=\frac{e}{3}\lim_{x\to 0}\frac{e^{t} -1}{t}\cdot\frac{t}{x}\\ &=\frac{e}{3}\lim_{x\to 0}\frac{t}{x}\\ &=\frac{e}{3}\lim_{x\to 0}\frac{\csc x\log(1+\sin x)-1-x}{x}\\ &=\frac{e}{3}\left(\lim_{x\to 0}\frac{\log(1+\sin x)-\sin x}{x\sin x}-1\right)\\ &=\frac{e}{3}\left(\lim_{x\to 0}\frac{\log(1+\sin x)-\sin x}{\sin^{2} x}\cdot\frac{\sin x}{x}-1\right)\\ &=\frac{e}{3}\left(\lim_{z\to 0}\frac{\log(1+z)-z}{z^{2}}-1\right)\\ &=\frac{e}{3}\left(\lim_{z\to 0}\dfrac{\left(z - \dfrac{z^{2}}{2} + \cdots\right)-z}{z^{2}}-1\right)\\ &=\frac{e}{3}\cdot\frac{-3}{2}=-\frac{e}{2}\end{aligned}$$ In the above derivation we have $z = \sin x$ and $$\begin{aligned}t&=\csc x\log(1+\sin x) -1-x\\ &= \frac{\log(1 + \sin x)}{\sin x} - 1 - x\\ &= \frac{\log(1 + z)}{z} - 1 - x\end{aligned}$$ so that both $t$ and $z$ tend to $0$ as $x\to 0$.

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  • $\begingroup$ Very nice solution ! Cheers :) $\endgroup$ – Claude Leibovici Jun 23 '14 at 15:57
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Hint

Taylor expansion built at $x=0$ leads to a solution since $$\csc(x)=\frac{1}{x}+\frac{x}{6}+O\left(x^2\right)$$ $$(1+\sin x)^{\csc x}=e-\frac{e x}{2}+O\left(x^2\right)$$ $$e^{x+1}=e+e x+O\left(x^2\right)$$ $$(1+\sin x)^{\csc x}-e^{x+1}=-\frac{3 e x}{2}+O\left(x^2\right)$$ $$\sin(3x)=3 x+O\left(x^2\right)$$ and then the limit equal to $-\frac{e}{2}$

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    $\begingroup$ And, for consistence, we should prove all these expansions without any use of De l'Hospital :-) $\endgroup$ – Siminore Jun 23 '14 at 11:15
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    $\begingroup$ While this is much shorter than my answer, finding Taylor series for $(1+\sin x)^{\csc x}$ is bit of a challenge for beginners. +1 for finding such short route to the answer. $\endgroup$ – Paramanand Singh Jun 23 '14 at 16:00

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