# intermediate limit - L'Hopital's rule?

• Nov 28th 2012, 01:00 AM
Erka
intermediate limit - L'Hopital's rule?
Hi, can anyone explain how

lim x->0 (x+( (e^(2x)) - (e^(3x)) / 3x+( e^(2x) )- (e^(5x)))

Can be 5/21?
When I try using L'Hopital I get -4/-18.
• Nov 28th 2012, 01:22 AM
MarkFL
Re: intermediate limit - L'Hopital's rule?
I assume you mean:

$L=\lim_{x\to0}\frac{x+e^{2x}-e^{3x}}{3x+e^{2x}-e^{5x}}$

We see we have the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$L=\lim_{x\to0}\frac{1+2e^{2x}-3e^{3x}}{3+2e^{2x}-5e^{5x}}$

We still have the indeterminate form 0/0, so application of L'Hôpital's rule again gives:

$L=\lim_{x\to0}\frac{4e^{2x}-9e^{3x}}{4e^{2x}-25e^{5x}}=\frac{4-9}{4-25}=\frac{5}{21}$
• Nov 28th 2012, 01:33 AM
Erka
Re: intermediate limit - L'Hopital's rule?
Thanks a lot!
• Nov 28th 2012, 01:37 AM
MarkFL
Re: intermediate limit - L'Hopital's rule?
Glad to help and welcome to the forums! :D