Question

Use L'Hopitals Rule to Calculate:

Lim (tan(2x))^x
X ->O^+

Answer 1

so far i have

Answer 2

i spose thats a good start

Answer 3

tan = sin/cos right?

Answer 4

not sure if that helps with the ^x; might have to do some ln to it

Answer 5

y = (tan(2x))^x

ln(y) = x ln(tan(2x))

y'/y = x' ln(tan(2x)) + x ln'(tan(2x))

y'/y = ln(tan(2x)) + x/tan(2x) *tan(2x)'

y'/y = ln(tan(2x)) + 2x sec^2(2x)/tan(2x)

not really sure how that would help tho lol

Answer 6

1
---
cs

2x sec(x)csc(x) ... just playing

Answer 7

\[\lim_{x\to 0}x\ln(\tan(2x))\] is in the form \(0\times -\infty\) standard trick is to write it as a fraction as
\[\frac{\ln(\tan(2x))}{\frac{1}{x}}\] and use l'hopital

Answer 8

first time you get
\[\frac{-2x^2\cos(2x)}{\sin(2x)}\] which is still \(\frac{0}{0}\) so repeat the process. second time you get 0

Answer 9

therefore your limit is
\[e^0=1\]

Answer 10

thanks! :)

Answer 11

yw
i left the algebra for you

Answer 12

ok cool :)