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.
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:
$\displaystyle L=\lim_{x\to0}\frac{4e^{2x}-9e^{3x}}{4e^{2x}-25e^{5x}}=\frac{4-9}{4-25}=\frac{5}{21}$