# Thread: problem's w/ l'hopital's rule

1. ## problem's w/ l'hopital's rule

I have dificulty solving these problems:

the first is:
lim x -> 0 (from the left) ((3x-2)/(3x+2))^x

I wasn't sure, could I just plug in 0 and get -1 instead of applying l'hopital's rule?

the second is:
lim x -> 1 (from the left) (x-1) (tan((pi * x)/2) )

(tan((pi * x)/2) ) / (1/(x-1)
since there is no multiplication rule. Took the derivative...

((sec(pi*x / 2)^2 * (pi/2)) / (-1/(x-1)^2)

2. I think the first one is indeterminate type $\displaystyle \infty ^0$. If lim f is of type $\displaystyle 0^0, \infty^0$or $\displaystyle 1^{\infty}$, then $\displaystyle lim(ln f)$ will be of the form $\displaystyle 0 . \infty$.

3. Originally Posted by Roam
I think the first one is indeterminate type $\displaystyle \infty ^0$. If lim f is of type $\displaystyle 0^0, \infty^0$or $\displaystyle 1^{\infty}$, then $\displaystyle lim(ln f)$ will be of the form $\displaystyle 0 . \infty$.
The limit of the first one is actually of the form $\displaystyle (-1)^{0^+}$ so l'hopitals rule does not apply. The function is discontinuous at x=0 and the limit doesn't exist.

4. Thanks makaan, that furthered my believe in my answer.

in #2,i tried the prob. again...
i did the l'hopital's (LH) rule and ( rearranged because of a lack of multiplication rule) and got,
(sec(pi * x/2)^2 * (pi / 2))/ ( -1 / (x - 1)^2) ... which is undef.

i did it again, this i rearranged ((x-1)^2 (pi/2))/ (-1 / sec(pi*x /2) and the denominator turn into -cos^2 and got...

(2(x-1)(pi/2) / (sin (pi*x / 2))^2 , i plugged in the 1 and got 0/1 = 0. is this right?