1. ## Hyperbolic Limit

Find the value of $\displaystyle\lim_{x\to{0}}\frac{sinh\:x-sin\:x}{sin^3\:x}$. Any method is fine. (obviously the fastest is best...)

Thanks!

2. Originally Posted by polymerase
Find the value of $\displaystyle\lim_{x\to{0}}\frac{sinh\:x-sin\:x}{sin^3\:x}$. Any method is fine. (obviously the fastest is best...)

Thanks!
Hello,

the only way I would do this problem is using the rule de l'Hôpital 3 times:

$\lim_{x\to{0}}\frac{\sinh(x)-\sin(x)}{\sin^3(x)} = \lim_{x\to{0}}\frac{\frac{e^x}{2}+\frac{e^{-x}}{2}-\cos(x)}{3\sin^2(x) \cdot \cos(x)} =$ $\lim_{x\to{0}}\frac{\frac{e^x}{2}-\frac{e^{-x}}{2}+\sin(x)}{9\sin(x) \cdot \cos^2(x)-3\sin(x)} =$ $\lim_{x\to{0}}\frac{\frac{e^x}{2}+\frac{e^{-x}}{2}+\cos(x)}{27 \cos^3(x) -21\cos(x)} = \frac26 = \frac13$

3. Originally Posted by polymerase
Find the value of $\displaystyle\lim_{x\to{0}}\frac{sinh\:x-sin\:x}{sin^3\:x}$. Any method is fine. (obviously the fastest is best...)

Thanks!
I don't know if this is more or less work than earboth's method. (I guess it depends on how many Maclaurin series you know.)

$sin(x) \approx x - \frac{x^3}{6}$

$sinh(x) \approx x + \frac{x^3}{6}$

So
$\lim_{x \to 0} \frac{sinh(x) - sin(x)}{sin^3(x)} \approx \lim_{x \to 0} \frac{ \left ( x + \frac{x^3}{6} \right ) - \left ( x - \frac{x^3}{6} \right )}{(x)^3}$
(We only need the first term in the sine expansion since we are approximating this to third order in x. Noticing this saves a lot of time since we don't have to expand the cubic.)

$= \lim_{x \to 0}\frac{ \frac{x^3}{3} }{x^3} = \frac{1}{3}$

-Dan

4. Originally Posted by topsquark
I don't know if this is more or less work than earboth's method. (I guess it depends on how many Maclaurin series you know.)

$sin(x) \approx x - \frac{x^3}{6}$

$sinh(x) \approx x + \frac{x^3}{6}$

So
$\lim_{x \to 0} \frac{sinh(x) - sin(x)}{sin^3(x)} \approx \lim_{x \to 0} \frac{ \left ( x + \frac{x^3}{6} \right ) - \left ( x - \frac{x^3}{6} \right )}{(x)^3}$
(We only need the first term in the sine expansion since we are approximating this to third order in x. Noticing this saves a lot of time since we don't have to expand the cubic.)

$= \lim_{x \to 0}\frac{ \frac{x^3}{3} }{x^3} = \frac{1}{3}$

-Dan
Firstly, thanks
Second, do you know a place where i can get a list of the Maclaurin series?

Thanks again

5. Second, do you know a place where i can get a list of the Maclaurin series?
this might be what you are looking for