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.

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- Nov 28th 2012, 12:00 AMErkaintermediate 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, 12:22 AMMarkFLRe: intermediate limit - L'Hopital's rule?
I assume you mean:

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

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

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

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

$\displaystyle L=\lim_{x\to0}\frac{4e^{2x}-9e^{3x}}{4e^{2x}-25e^{5x}}=\frac{4-9}{4-25}=\frac{5}{21}$ - Nov 28th 2012, 12:33 AMErkaRe: intermediate limit - L'Hopital's rule?
Thanks a lot!

- Nov 28th 2012, 12:37 AMMarkFLRe: intermediate limit - L'Hopital's rule?
Glad to help and welcome to the forums! :D