Question

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

Answer 1

L'hospital's rule: indeterminate form of 0/0

Does this help?

Answer 2

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

Answer 3

So derive the top and bottom seperately

Answer 4

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

Answer 5

before I can take the derivative

Answer 6

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

Answer 7

the minus 1 makes it go to 0

Answer 8

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

Answer 9

Medal pleaser ^.^

Answer 10

Ohh ok thanks

Answer 11

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

Answer 12

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

Answer 13

\[y = (1+h)^{\ln 1+h} \rightarrow \ln y=\ln (1+h)^{\ln 1+h}\]
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)
\[\frac{dy}{dx} = 2\ln (1+h)*(1+h)^{\ln (1+h) -1}\]