Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to find $\displaystyle\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ using Taylor polynomials.

Here's what I've done so far:

  1. $e^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$
  2. $\sin x = x - \frac{1}{6} x^3 + o(x^4)$
  3. $e^{\sin x} = 1 + x - \frac{1}{6} x^3 + o(x^4)$
  4. $\sin 2x = 2x + o(x^2)$
  5. $\sin^3 2x = 8x^3 + o(x^6)$

Therefore I can rewrite my limit as: $\displaystyle\lim_{x \rightarrow 0} \frac{-\frac{1}{3} x^3 - \frac{1}{2} x^2 + o(x^3)}{8x^3 + o(x^6)}$

In this form, the limit appers to be $-\frac{1}{24}$, but the correct result is $-\frac{1}{48}$.

Could you tell me what I'm doing wrong?

share|improve this question
1  
The first equation is wrong. correct one is $\mathbb{e}^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{\color{red}{6}} x^3 + o(x^3)$ –  Shane Chern Jan 28 '13 at 12:46
    
Argh, what a silly mistake! Fixed that and an another copy&paste error. –  hey hey Jan 28 '13 at 12:50
    
You forgot to add $\frac{1}{2}\sin^2 x=\frac{1}{2}x^2+o(x^3)$ and $\frac{1}{6}\sin^3 x=\frac{1}{6}x^3+o(x^3)$... –  Shane Chern Jan 28 '13 at 12:59
add comment

2 Answers

up vote 1 down vote accepted

$\mathbb{e}^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$ and $\sin x=x-\frac{1}{6}x^3+o(x^3)$
$\Rightarrow \mathbb{e}^{\sin x}=1+x-\frac{1}{6}x^3+\frac{1}{2}x^2+\frac{1}{6}x^3+o(x^3)=1+x+\frac{1}{2}x^2+o(x^3)$
$\Rightarrow \mathbb{e}^{\sin x}-\mathbb{e}^x=-\frac{1}{6}x^3$.

share|improve this answer
add comment

Let's rewrite the limit as $$\displaystyle\lim_{x \rightarrow 0} e^x\times\lim_{x \rightarrow 0}\frac{e^{\sin x-x} - 1}{\sin x-x}\times\lim_{x \rightarrow 0}\left(\frac{2x}{\sin 2x}\right)^3\times \lim_{x \rightarrow 0}\frac{\sin x-x}{8x^3}=\lim_{x \rightarrow 0}\frac{x-x^3/6+O(x^5)-x}{8x^3}=-\frac{1}{48}$$

Q.E.D.

share|improve this answer
add comment

Your Answer

 
discard

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.