# Math Help - Don't understand some part of l'hopital's rule

1. ## Don't understand some part of l'hopital's rule

6. c) Use l'Hopital's rule to evaluate lim x->0 e^x - e^-x -2x / x - sin x

step 1 = e^x + e^-x -2 / 1 - cos x

step 2 = e^x - e^-x / sin x

step 3 = e^x + e^-x / cos x

step 4 = above is = to 2/1

I do not understand how 1-cosx becomes sin x, I thought it would be -sinx. Secondly, I do not understand how step 3 is equal to 2/1.

2. Originally Posted by thekrown
6. c) Use l'Hopital's rule to evaluate lim x->0 e^x - e^-x -2x / x - sin x

step 1 = e^x + e^-x -2 / 1 - cos x

step 2 = e^x - e^-x / sin x

step 3 = e^x + e^-x / cos x

step 4 = above is = to 2/1

I do not understand how 1-cosx becomes sin x, I thought it would be -sinx. Secondly, I do not understand how step 3 is equal to 2/1.
Please either use LaTeX or correct use of brackets. What you have posted is very difficult to read.

$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}}$

using L'Hospital's Rule...

First check that it is indeterminate.

You should see that it tends to

$\frac{e^0 - e^0 - 2(0)}{0 - \sin{0}}$

$= \frac{0}{0}$.

So you can use L'Hospital.

Now, take the derivative of the top, and the derivative of the bottom.

You will have

$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x} - 2}{1 - \cos{x}}$.

Now see what happens when you let it $\to 0$.

You should get

$\frac{e^0 + e^0 - 2}{1 - \cos{0}}$

$= \frac{0}{0}$.

It is still indeterminate, so use L'Hospital again.

$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x} - 2}{1 - \cos{x}} = \lim_{x \to 0}\frac{e^x - e^{-x}}{\sin{x}}$

(note that since the derivative of $\cos{x} = -\sin{x}$ then the derivative of $-\cos{x} = \sin{x}$.)

$= \frac{e^0 - e^0}{\sin{0}}$

$= \frac{0}{0}$.

Use L'Hospital again...

$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x} - 2}{1 - \cos{x}} = \lim_{x \to 0}\frac{e^x - e^{-x}}{\sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x}}{\cos{x}}$

$= \frac{e^0 + e^0}{\cos{0}}$

$= \frac{1 + 1}{1}$

$= 2$.