# Homework Help: Limit as x->0+

1. Jan 6, 2010

### Jules18

Limit as x-->0+

1. The problem statement, all variables and given/known data
lim (Cosx)1/(x2)
x-->0+

3. The attempt at a solution

I figured 1/x2 would approach infinity, which and Cos0=1, so the function would approach 1$$\infty$$ which is $$\infty$$, but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks

2. Jan 6, 2010

### Staff: Mentor

Re: Limit as x-->0+

$$[1^{\infty}]$$ is an indeterminate form, which means that a limit of this type can come out as just about anything.

I would let y = (cos x)1/x2, and then take ln of both sides. Then take the limit of that as x --> 0+. In your notes or text there might be a similar example that you can use as a template.

3. Jan 6, 2010

### Jules18

Re: Limit as x-->0+

Really? I always thought the limit of 1/(x2) would be infinite when x approaches 0.

4. Jan 6, 2010

### Staff: Mentor

Re: Limit as x-->0+

This is not true at all. If 1/x2 approaches 0, then x approaches either positive or negative infinity, not zero.

5. Jan 6, 2010

### vela

Staff Emeritus
Re: Limit as x-->0+

I'm just guessing, but I think the answer comes out to exp(-1/2).

Take the log of the function to get the variable out of the exponent and find the limit of the log. The answer will be e to the limit you find.

6. Jan 6, 2010

### yungman

Re: Limit as x-->0+

Sorry!!! My mind is not working today, still in the Christmas spirit!!!! Ignor me totally and I'll stay out of this forum today!!! It's not me, it's my finger......Sorry!!!

7. Jan 7, 2010

### HallsofIvy

Re: Limit as x-->0+

$1^\infty$ is NOT 1, it is "indeterminate". If $y= (cos(x))^{1/x^2}$, then $ln(y)= (ln cos(x))/x^2$ which is indeterminate of the form "0/0". Use L'Hopital's rule on that.

8. Jan 7, 2010

### Gregg

Re: Limit as x-->0+

$$\ln (y) = \displaystyle \lim_{x\to 0^+} \frac{ln (\cos x)}{x^2}$$

$$\ln (y) = \lim_{x\to 0^+} \frac{-\sin x}{2x\cos x} = \lim_{x\to 0^+} \frac{-x+{x^3\over 3!}-\cdots}{2x(1-{x^2\over 2!}\cdots)} =-\frac{1}{2}$$

$$y=\frac{1}{\sqrt{e}}$$

Last edited: Jan 7, 2010
9. Jan 7, 2010

### Dick

Re: Limit as x-->0+

Nah. The limit from the left and the right are the same. But your solution has a problem. The numerator is supposed to be log(cos(x))/x^2, not cos(x)/x^2. The latter is infinite.

10. Jan 7, 2010

### vela

Staff Emeritus
Re: Limit as x-->0+

11. Jan 7, 2010

### Gregg

Re: Limit as x-->0+

Oh yeah, I think it's OK now.

12. Jan 8, 2010

### Jules18

Re: Limit as x-->0+

I understand up to = $$\lim_{x\to 0^+} \frac{-x+{x^3\over 3!}-\cdots}{2x(1-{x^2\over 2!}\cdots)}$$.

I really lost you on that one. Why would that be equal to lim -Sinx/(2xCosx) ??

13. Jan 8, 2010

### Staff: Mentor

Re: Limit as x-->0+

Gregg replaced -sin x and cos x by their Maclaurin series. This might be an approach that is more advanced than you are prepared for at this time.

14. Jan 11, 2010

### Jules18

Re: Limit as x-->0+

hm okay well I was given this question on an exam, so I figure my prof had a different approach in mind. Is there a simpler way?

15. Jan 11, 2010

### vela

Staff Emeritus
Re: Limit as x-->0+

After taking the log, just use the hospital rule twice.