1. ## l'hopital Question

Hiii :3

I'm stuck on this question:

lim (x->1) (2-x)^ tan(π/2)x

I did the ln thing but now i'm stuck what do I do next?
Plug it in?

Thank youuus :]

2. Originally Posted by TeriyakiDonnQ
Hiii :3

I'm stuck on this question:

lim (x->1) (2-x)^ tan(π/2)x

I did the ln thing but now i'm stuck what do I do next?
Plug it in?

Thank youuus :]
Seeing as you've been told to use L'Hospital's Rule, it would make sense to use L'Hospital's Rule, wouldn't it?

You are correct that you should apply the exponential-logarithmic transformation...

$\displaystyle \lim_{x \to 1} (2 - x)^{\tan{\frac{\pi x}{2}}} = \lim_{x \to 1}e^{\ln{(2 - x)^{\tan{\frac{\pi x}{2}}}}}$

$\displaystyle = \lim_{x \to 1}e^{\tan{\left(\frac{\pi x}{2}\right)}\ln{(2 - x)}}$

$\displaystyle = \lim_{x \to 1}e^{\frac{\sin{\left(\frac{\pi x}{2}\right)}\ln{(2 - x)}}{\cos{\left(\frac{\pi x}{2}\right)}}}$

$\displaystyle = e^{\lim_{x \to 1}\frac{\sin{\left(\frac{\pi x}{2}\right)}\ln{(2 - x)}}{\cos{\left(\frac{\pi x}{2}\right)}}}$.

This limit is now of the form $\displaystyle \frac{0}{0}$, so now you can use L'Hospital's Rule.