How do I solve this difficult L'Hopital problem using fractions and derivatives?

[Frillth]

I got the following problem in my math class:

http://img81.imageshack.us/img81/7508/limitdv4.jpg

I know that I'm supposed to use L'Hopital's rule, but I have 2 problems. First of all, I don't know how to make that into a fraction, besides putting it all over 1 or making tan negative and putting it under 1. Also, I have no idea how to even find the derivative of that mess.

I know that the answer is e^(2/pi), but I'd like to know how to get it.

Can anybody help me, please?

 

[Mindscrape]

Right now it is in indeterminate form, so first thing you will have to do is figure out how to fix that since L'Hopital only works for (inf/inf) and (0/0). You might try making some function i.e. L(x) equal to that limit. Then perhaps use log rules and such. I haven't actually worked it out to see if that would work since that is your job, but play around and you will get it.

 

[VietDao29]

I got the following problem in my math class:

http://img81.imageshack.us/img81/7508/limitdv4.jpg

I know that I'm supposed to use L'Hopital's rule, but I have 2 problems. First of all, I don't know how to make that into a fraction, besides putting it all over 1 or making tan negative and putting it under 1. Also, I have no idea how to even find the derivative of that mess.

I know that the answer is e^(2/pi), but I'd like to know how to get it.

Can anybody help me, please?

Ok, you can solve the problem like this:
Let: $$y = \lim_{x \rightarrow 1} (2 - x) ^ {\tan \left( \frac{\pi}{2} x \right)}$$
Now, by taking log of both sides, and using the fact that the function ln(x) is continuous for all x > 0, we have:
$$\ln y = \ln \left( \lim_{x \rightarrow 1} (2 - x) ^ {\tan \left( \frac{\pi}{2} x \right)} \right) = \lim_{x \rightarrow 1} \left[ \ln \left( (2 - x) ^ {\tan \left( \frac{\pi}{2} x \right)} \right) \right] = \lim_{x \rightarrow 1} \left[ \tan \left( \frac{\pi}{2} x \right) \ln (2 - x) \right]$$
The limit in the right is in the Indeterminate form $$0 \times \infty$$
Now, for it to have the form $$\frac{0}{0} \quad \mbox{or} \quad \frac{\infty}{\infty}$$, we can divide both numerator, and denominator by ln(2 - x), or tan(pi x / 2), like this:
$$\ln y = \lim_{x \rightarrow 1} \left[ \tan \left( \frac{\pi}{2} x \right) \ln (2 - x) \right] = \lim_{x \rightarrow 1} \frac{\tan \left( \frac{\pi}{2} x \right)}{\frac{1}{\ln (2 - x)}}$$
Can you go from here?
After evaluating the limit in the right, can you find y, i.e the limit of the original problem? :)

 

[dextercioby]

1. Make the sub

$$x=y+1$$

2. Then

$$\lim_{y\rightarrow 0^{+}} (1-y)^{\frac{1}{y}\cdot \[y\tan\left(\frac{\pi}{2}(1+y)\right)}$$

U can write it as e^(-lim...) and you may use (if you wish) l'H^opital's rule to evaluate the limit y/cos(pi/2(1+y)) as y tends to zero plus.

Daniel.