## anonymous 5 years ago how do you evaluate the lim h->0 [(1+h)^ln(l+h) -1]/ h

1. anonymous

L'hospital's rule: indeterminate form of 0/0 Does this help?

2. dumbcow

i agree, differentiate top and bottom and i think limit is zero

3. anonymous

So derive the top and bottom seperately

4. anonymous

Does the (1+h)^(ln(1=h) go to 0 already or do I have to rearrange ?

5. anonymous

before I can take the derivative

6. anonymous

or does it go to one making it 1-1 so i have a 0/0 function

7. anonymous

the minus 1 makes it go to 0

8. anonymous

so (1+h)^(ln(1=h)= 1 but there is a minus 1 in the numerater so its 0

9. anonymous

10. anonymous

Ohh ok thanks

11. anonymous

so you should show your work, and u'll get 0

12. anonymous

Im slightly confused on how to take the derivative of (1+h)^ln(1+h)

13. dumbcow

$y = (1+h)^{\ln 1+h} \rightarrow \ln y=\ln (1+h)^{\ln 1+h}$ use log rule log x^n = n*log x $\ln y=(\ln (1+h))^{2}$ now differentiate $\frac{1}{y}\frac{dy}{dx} = \frac{2}{1+h} \ln (1+h)$ multiply by y on both sides where y=(1+h)^ln(1+h) $\frac{dy}{dx} = 2\ln (1+h)*(1+h)^{\ln (1+h) -1}$