# Finding Limit With L'Hospital's Rule

• Nov 22nd 2011, 10:06 PM
NeoSonata
Finding Limit With L'Hospital's Rule
lim x->1 (2-x)^(tan(pi(x)/2))

I plugged it in and got 1^infinity (indeterminate form) because tan (pi(1)/2) is 1/0 or infinity

so:

lny = (tan(pi(x)/2)) * ln(2-x)

= ln(2-x) / cot (pi(x)/2)

L'H = (-1/2-x) / csc^2(pi(x)/2) ( -pi/2)

= -1/ -pi/2
= 2/pi

lim x->1 y = e^(2/pi)

?

Where did I go wrong?

Thanks
• Nov 22nd 2011, 10:21 PM
Soroban
Re: Finding Limit With L'Hospital's Rule
Hello, NeoSonata!

• Nov 22nd 2011, 10:29 PM
NeoSonata
Re: Finding Limit With L'Hospital's Rule
I checked the answer using wolfram alpha

it got:
lim x->1 (2-x)^(tan(pi(x)/2)) = 1

the site does not show any work for this problem though.
• Nov 23rd 2011, 01:39 AM
Prove It
Re: Finding Limit With L'Hospital's Rule
Quote:

Originally Posted by NeoSonata
lim x->1 (2-x)^(tan(pi(x)/2))

I plugged it in and got 1^infinity (indeterminate form) because tan (pi(1)/2) is 1/0 or infinity

so:

lny = (tan(pi(x)/2)) * ln(2-x)

= ln(2-x) / cot (pi(x)/2)

L'H = (-1/2-x) / csc^2(pi(x)/2) ( -pi/2)

= -1/ -pi/2
= 2/pi

lim x->1 y = e^(2/pi)

?

Where did I go wrong?

Thanks

You must have entered it into Wolfram wrongly, since the limit is $\displaystyle e^{\frac{2}{\pi}}$.

limit of &#91;&#40;2-x&#41;&#94;&#40;Tan&#91;Pi &#42; x &#47; 2&#93;&#41;&#93; as x to 1 - Wolfram|Alpha

\displaystyle \begin{align*} \lim_{ x \to 1} (2 - x)^{\tan{\frac{\pi x}{2}}} &= \lim_{x \to 1}e^{\ln{\left[(2 - x)^{\tan{\frac{\pi x}{2}}}\right]}} \\ &= \lim_{x \to 1}e^{\tan{\left(\frac{\pi x}{2}\right)}\ln{\left(2-x\right)}} \\ &= e^{\lim_{x \to 1}\tan{\left(\frac{\pi x}{2}\right)}\ln{\left(2-x\right)}} \\ &= e^{\lim_{x \to 1}\frac{\sin{\left(\frac{\pi x}{2}\right)}\ln{\left(2-x\right)}}{\cos{\left(\frac{\pi x}{2}\right)}}} \\ &= e^{\lim_{x \to 1}\frac{\frac{\pi}{2}\cos{\left(\frac{\pi x}{2}\right)}\ln{(2-x)} - \frac{\sin{\left(\frac{\pi x}{2}\right)}}{2-x}}{-\frac{\pi}{2}\sin{\left(\frac{\pi x}{2}\right)}}}\textrm{ by L'Hospital's Rule} \\ &= e^{\frac{\frac{\pi}{2} \cos{ \left( \frac{\pi}{2} \right) } \ln { (2-1) } - \frac{\sin{\frac{\pi}{2}}}{2-1}}{-\frac{\pi}{2}\sin{\frac{\pi}{2}}}} \\ &= e^{\frac{\frac{\pi}{2}\cdot 0 \cdot 0 - \frac{1}{1}}{-\frac{\pi}{2} \cdot 1}} \\ &= e^{\frac{2}{\pi}} \end{align*}
• Nov 30th 2011, 03:30 PM
CalBear12
Re: Finding Limit With L'Hospital's Rule
Just take the derivative of the numerator and denominator if the first result is indeterminant.