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 can do this limit with a symbolic calculator and get the result.

$$\lim_{x=0} \left[ \frac{1}{x^2}\left(\frac{\sinh x}{x} - 1\right) \right] = \frac{1}{6}$$

But how would I do it by hand, and show why it is so. I know that $\lim_{x=0}\frac{\sinh(x)}{x}=1$ but that does not help here.

This is not homework, and it is related to the deflection of axially loaded beams.

share|improve this question
    
Writing it as $(\sinh(x) - x)/ x^3$ and applying L'hopital three times will work, as will using the Taylor series for $\sinh(x)$. –  Zarrax Jul 18 '12 at 19:59
add comment

2 Answers

up vote 5 down vote accepted

HINT: $$\sinh (x) = \dfrac{e^x - e^{-x}}2 = x + \dfrac{x^3}6 + \mathcal{O}(x^5)$$

share|improve this answer
1  
Aha! Thanks. Did you even have to think about this at all? That was a quick answer. –  ja72 Jul 18 '12 at 19:56
add comment

You can also simply use l’Hospital’s rule:

$$\begin{align*} \lim_{x\to 0} \left[ \frac{1}{x^2}\left(\frac{\sinh x}{x} - 1\right) \right]&=\lim_{x\to 0}\frac{\sinh x-x}{x^3}\\ &=\lim_{x\to 0}\frac{\cosh x-1}{3x^2}\\ &=\lim_{x\to 0}\frac{\sinh x}{6x}\\ &=\lim_{x\to 0}\frac{\cosh x}6\\ &=\frac16\;. \end{align*}$$

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.