Limit answer check

Question

So I have a limit problem as $x$ approaches zero.

My function is : $\frac{\sinh x-\sin x}{x^3}$

I use L'Hopital three times, then I get the limit to be $0$.

However, the answer is supposed to be $0.33333\dots$.

What did I do wrong here?

It was supposed to be sinhx-sinx / x^3, but I did something wrong with the latex command... — user140878, Oct 17 '14 at 00:34
Did I edit it right, or did you mean $\frac{\sinh x-\sin x}{x^3}$? — Edward Jiang, Oct 17 '14 at 00:35

score 1 · Answer 1 · answered Oct 17 '14 at 01:18

You differentiated $\sinh(x)$ wrong would be my guess. So, let's compute using the definition of $\sinh(x)=(e^x-e^{-x}/2)$.

Then you want limit of $\dfrac{(e^x-e^{-x})-2\sin(x)}{2x^3}$ as $x \rightarrow 0$.

Use L-Hospital once:

Then we have to find limit of : $\dfrac{(e^x+e^{-x})-2\cos(x)}{6x^2}$.

Looks like another use of L-Hospital is due. So here we go:

$\dfrac{(e^x-e^{-x})+2\sin(x)}{12x}$.

Another use of L-Hospital gives:

$\dfrac{(e^x+e^{-x})+2\cos(x)}{12}$- which has limit $4/12=1/3=0.333\cdots$ as $x \rightarrow 0$.

1 Answers1