# 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.