# Homework Help: Help w/ 1 more limit

1. Dec 16, 2004

### SomeRandomGuy

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

I took the ln and got tan(x)ln(x), then made it ln(x)/(1/tan(x)) which = ln(x)/(cot(x)) and I could use L' Hospital's rule. I got (1/x)/(csc^2(x)) and made that (1/x)/(1/sin^2(x)). I then made that sin^2(x)/x and used L' Hospital's rule again to get the lim as x approaches 0 from the right of 2cos(x) which = 2. Am I right or wrong? :/

2. Dec 16, 2004

### Hurkyl

Staff Emeritus
You've made at least one mistake.

3. Dec 16, 2004

### Hurkyl

Staff Emeritus
0^0 is not 1. :tongue2: It's an indeterminate form.

4. Dec 16, 2004

### Euclid

Write $$L=\lim_{x\rightarrow 0} x^{\tan x}$$.
Then $$\log L =\lim_{x\rightarrow 0} \tan x\log x = \lim_{x\rightarrow 0} \frac{\tan x}{\log x}= 0$$
So L = e^0 = 1. There is no use for L'Hopitals rule here.

5. Dec 16, 2004

### Cantari

0/0 is another indeterminate form. When you took the derivative of sin^2 x, you messed up (as well as forgot a negative sign which doesnt effect it). So instead of your top going to 2 it should go to 0.

Last edited: Dec 16, 2004
6. Dec 16, 2004

### dextercioby

U've written something wrong.
Limit and logaritm commute.Use this property to show that:
$$\lim_{x\rightarrow 0} \ln(x^{\tan x}) = \lim_{x\rightarrow 0}\tan x\ln x$$

I hope u can show that the logaritm of the limit is zero,and hence the initial limit is 1.

Daniel.

7. Dec 16, 2004

### dextercioby

Not to perform any differentiations on the sine squared use this trick
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} =1$$.

Daniel.

8. Dec 16, 2004

### Cantari

Also I am not sure that you know that when you take the ln of this, you need to take the ln of the other side of the equals sign. So just assume that it equals y, then the answer is ln y = 0, which would be 1.

9. Dec 16, 2004

### dextercioby

Which other side??Of course u apply logarithm on the both sides of the equation:
$$\lim_{x\rightarrow 0} x^{\tan x} = L$$
,where L is the limit and is the unknown.When u take logarithm,u should be stating:
$$\ln L=...=0$$
,from where u get your result...

Daniel.

10. Dec 16, 2004

### Cantari

... my comment was directed to the original poster as he said he got an answer of 2 and assumed that was the final answer. Which would not be the case even if 2 was what the right side came out to be due to the fact he took the log of the problem in the first step.