Question

Evaluate using l'Hospital's Rule:
lim x-> 1 (3-2x)^(1/x-1)

Answer 1

\[\lim\limits_{x\rightarrow 1}\quad(3-2x)^{\frac 1{x-1}}\]

Answer 2

/?

Answer 3

Yup, that's the function.

Answer 4

I honestly don't know, I don't have the answer to the problem on me.

Answer 5

Oh okay. So like you said the answer is, 1^(infinity) = 1?

Answer 6

To evaluate this one, we note that we have an indeterminate form of the type 1^infinity. To evaluate this we take the natural log of both sides:\[\ln y=\frac{1}{x-1}\ln (3-2x)\]Now we have an indeterminate form of type 0/0 when x===>1. So,\[\lim_{x \rightarrow 1}\ln y=\lim_{x \rightarrow 1}\frac{-2}{(3-2x)}=-2\]So,\[\lim_{x \rightarrow 1}y(x)=\lim_{x \rightarrow 1}e^{\ln y(x)}=e^{-2}\]

Answer 7

This makes so much more sense! Thank you x infinity! :)

Answer 8

:)