Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
Please see the editing. I think there is a mistake here. Did you mean $e^x$ or $e^{x+1}$? – Tunk-Fey Jun 23 '14 at 10:25
@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. – Joffysloffy Jun 23 '14 at 10:26
@Joffysloffy Sorry, I thought it was equal to $1$. – Tunk-Fey Jun 23 '14 at 10:29
@Tunk-Fey, That's okay :). Note that it is equivalent to the limit definition of $e$ by using substitution of limits. – Joffysloffy Jun 23 '14 at 10:30
up vote 6 down vote accepted

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 $x\to 0$.

share|cite|improve this answer
Very nice solution ! Cheers :) – Claude Leibovici Jun 23 '14 at 15:57


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 $-\frac{e}{2}$

share|cite|improve this answer
And, for consistence, we should prove all these expansions without any use of De l'Hospital :-) – Siminore Jun 23 '14 at 11:15
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. – Paramanand Singh Jun 23 '14 at 16:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.