# L' Hospital's Rule Application

1. Dec 10, 2006

### carmen77

1. The problem statement, all variables and given/known data
$$\lim_{x\rightarrow 0} \frac{xe^2x+xe^x-2e^2x+2e^x}{(e^x-1)^3}$$

2. Relevant equations
L' Hospital's Rule
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}= \lim_{x\rightarrow\0} \frac{f'(x)}{g'(x)}$$

3. The attempt at a solution

Phew, I think my hold on the latex codes are solid now, thanks Status. Okay, here's what I've worked so far, and where I'm stuck at.

Just as Hosp. Rule says, I found the derivative of the numerator and denominator functions seperately by using the chain rule:

$$\lim_{x\rightarrow 0} \frac{e^2x(2x+1)+e^x(x+1)-4e^2x+2e^x}{(3e^x)(e^x-1)^2}$$

I noticed that the function was still indeterminate due to the denominator going to 0, so I found the derivatives of the numerator and denominator again;

But my result showed that the denominator went to 0 again because its dervivative is:

$$3e^x(e^x-1)(3e^x-1)$$

I must be doing something wrong.

Last edited: Dec 10, 2006
2. Dec 10, 2006

### StatusX

You'll eventually get there, but it'll take a few times going through the process. You can make the algebra alot easier if you make the substitution u=e^x (so that the limit is now as u->1).

3. Dec 10, 2006

### calcnd

I did it quickly, so my answer might be off, but keep doing L'Hopital's rule until you get a proper answer.

(I got $$\frac{1}{6}$$, by the way.

Basically, just keep differentiating until you get rid of the 1 in the $$e^{x}-1$$ bracket, as then you'll get a proper answer.

4. Dec 10, 2006

### carmen77

Jeez again, man our teacher is really making us grind out these calculations...well okay thank you fellas.