Indeterminate Form_Limits

1. Aug 4, 2013

Justabeginner

1. The problem statement, all variables and given/known data
Find the following limit:

lim x-> 0+ ((1/x)^(tan x))

2. Relevant equations

3. The attempt at a solution
This gives me the indeterminate form infinity raised to power zero.
After trying two methods I still end up with an indeterminate form.

e^ ln (1/x)^(tan x)= e^[ln (1/x) * (tan x)]
(e^u) * d/dx (ln (1/x) (tan x))
(1/x)^(tan x) * d/dx (ln (1/x) (tan x))
(1/x)^(tan x) * [(ln(1/x)sec^2(x)) + (1/x)(tan x))]
That gives me an indeterminate form.

My second method:
ln f(x)= (tan x) * (ln (1/x))
f(x)= e^[(tan x) * (ln (1/x))]
lim x->0+ (1/x)^(tan x)= e ^ [lim x->0+ ((tan x) * (ln (1/x)))]
Also, what I think is an indeterminate form.
I'm inclined to say the answer is one, but I know that infinity to the power zero is invalid, and indeterminate. I'm stumped. Help anyone? Thanks so much!

2. Aug 4, 2013

lurflurf

Be careful it is possible to make matters worse. I am not able to follow your work completely. Taking the first attempt we have
$$\lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{\tan(x)}= \lim_{x \rightarrow 0^+} \exp \, \log \left( \frac{1}{x} \right)^{\tan(x)}\\ \text{The exp can be moved out of the limit as it is continuous. We want \infty/\infty form like this} \\ = \exp \lim_{x \rightarrow 0^+} -\log(x)/\cot(x)\\$$
Apply l'Hôpital's rule. At each application of l'Hôpital's rule look to rearange. Multiple applications of l'Hôpital's without care can complicate matters. See if you can finish.

3. Aug 4, 2013

Justabeginner

Yes sir, that is what I have. Using the natural log to get rid of the indeterminate form of infinity to the power zero. Sorry it was unclear.

I tried using L'hopital's rule but I still get indeterminate forms of infinity/infinity.

(-1/x)/(-csc^2(x))
(1/(x^2)/(2 cot x* csc^2(x)) (L'Hopital's Rule)

4. Aug 4, 2013

lurflurf

Yes as I hinted you want to rearrange the equation at this point
$$(1/x)/\csc^2(x)=\sin^2(x)/x=x(\sin(x)/x)^2=(1/2)(1-\cos(2x))/x$$
Any of these rearrangements will work much better. Do you know the limit of sin(x)/x?

5. Aug 4, 2013

Justabeginner

Yes I believe that it is zero.
And I understand now how you used those trigonometric identities in the last steps.
So since I get a 0/0 indeterminate form in the last step again, I use L'Hopital's if I am not mistaken, and I have:

sin(2x)/1= sin(2x). Then when lim x->0+ I have sin(2x)= sin(0)= 0?

Thank you.

6. Aug 4, 2013

Justabeginner

I'm not sure if what I have here is correct. Can anyone please confirm?

7. Aug 4, 2013

lurflurf

Yes that is right. Remember the original integral was e^ new one so L=e^lim sin(2x)

8. Aug 4, 2013

Justabeginner

Yes, so e^0= 1. Thank you for all your help. I really appreciate it.