1. Problem using l'hopital rule

Hi all

I have encountered a problem when applying the l'hopital rule to the following limit; could anyone help to see where i went wrong?

lim x->infinity (1+e^x+2e^2x) / (x+e^x+e^2x) ------(1);

repeating application of the rule gives:

lim x->infinity (e^x+4e^2x) / (1+e^x+4e^2x);

which then gives

lim x->infinity (e^x+8e^2x) / (e^x+8e^2x) = 1;

However in the solution provided, e^-2x is multiplied to (1), giving:

lim x->infinity (e^-2x+e^-x+2) / (xe^-2x+e^-x+1) = 2/1 = 2;

which is different from my answer.

Any help is much appreciated

2. EDIT: l'hoptials rule does not apply to this function. here is how you solve it:
$\displaystyle\lim_{x\rightarrow \infty}\frac{1+e^x+2e^{2x}}{x+e^x+e^{2x}}$

$\displaystyle\lim_{x\rightarrow \infty}\frac{\frac{1}{e^{2x}}+\frac{1}{e^x}+2}{\fr ac{x}{e^{2x}}+\frac{1}{e^x}+1}$

$=\frac{2}{1}$

3. In your solution, even after second derivative, the problem is in indeterminate form. Because it is a fraction of infinity/infinity. So you have to remove e^x from the numerator and denominator. That is what you have seen in the solution.

4. Thanks for the replies; upon double checking i realised that i have differentiated one part wrongly.