Thread: L'Hospital's Rule

Ok new chapter. I think I under stand this stuff to a point. I have :

lim x -> 0 e^x -1 -x -(x^2/2)/ x^3 so eventually after taking the derivative of the top and bottom about 3 times i ended up with e^x/6 so the limit approaches infinite correct?

Then

lim x -> 0 (1+a/x)^bx, this one is slightly harder i can get the first derivative then i'm not sure what to do. So for the first derivative I have b(1+a/x)^bx-1*(a/x^2) then i'm lost. {edit} ok i just realized different form g(x) ln f(x) so if i do that i get b * ( -a/x^2) so -ab/x^2 would that be negative infinite then?

Originally Posted by UMStudent

Ok new chapter. I think I under stand this stuff to a point. I have :

lim x -> 0 e^x -1 -x -(x^2/2)/ x^3 so eventually after taking the derivative of the top and bottom about 3 times i ended up with e^x/6 so the limit approaches infinite correct?

Look right.

Then

lim x -> 0 (1+a/x)^bx, this one is slightly harder i can get the first derivative then i'm not sure what to do. So for the first derivative I have b(1+a/x)^bx-1*(a/x^2) then i'm lost. {edit} ok i just realized different form g(x) ln f(x) so if i do that i get b * ( -a/x^2) so -ab/x^2 would that be negative infinite then?

If a,b>0

(1+ax)^(bx)=exp(ln(1+ax)^(bx)) for x>0.
But then,
exp(bx*ln(1+ax))

Note that,

bx*ln(1+ax) = ln(1+ax)/(1/bx)
You can use L-Hopital Rule on this.

so then ln(1+ax)/(1/bx) the derivative using L'Hospital's rule is (-a/x^2)/(1+ax)/(b), then from there how do i solve the limit as x -> infinite?