Expert: Paul Klarreich - 3/15/2006

Question
Im completly lost on what to do here, i tried taking the ln of both sides, but it didnt seem right. So im completly lost on what to do. Can you please help?

Find the limit using L'Hospital's Rule.  If it doesn't apply consider an alternative method, and explain why L'Hospital's Rule doesn't apply.

lim(as x approaches 0 from the right) of (tan2x)^x

Thanks so much


Answer
Hi, Chevonne,

Subject:  L'Hospital's Rule
Question:  Im completly lost on what to do here, i tried taking the ln of both sides, but it didnt seem right. So im completly lost on what to do.

Can you please help?

Find the limit using L'Hospital's Rule. If it doesn't apply consider an alternative method, and explain why L'Hospital's Rule doesn't apply.

lim(as x approaches 0 from the right) of (tan2x)^x

Thanks so much
-----------------------------
Your instincts were OK.  For weird functions like (...)^x, logarithms are a good choice.

Let  y =  (tan 2x)^x

ln y =  x ln tan 2x
        ln tan 2x
    =  ----------  
            1/x

[See -- you MUST make it a fraction somehow, even if you have to do that.]

Now use l'Hospital's rule on that, differentiating top and bottom:

(1/tan 2x)(sec^2 2x)(2)
------------------------
     -1/x^2

Now the top is

2 sec^2 2x   2(1/cos^2 2x)         2
---------- = ------------- = --------------
 tan 2x     sin 2x/cos 2x   sin 2x cos 2x

So the function is:

   - 2x^2         - 4 x^2        - 4x^2
------------- = --------------- = ------
sin 2x cos 2x   2 sin 2x cos 2x   sin 4x

Now use the rule again on this:

 - 8x      - 2x
-------- = -------
4 cos 4x    cos 4x

But now as x -> 0, you get  0/1 = 0.

So  ln y --> 0, means  y -> 1