# Limit using L'hospital's rule

1. Jan 4, 2009

### sara_87

1. The problem statement, all variables and given/known data

Calculate the limit as x tends to 0:

(sinx/x)^(1/x^2)

2. Relevant equations

3. The attempt at a solution

I have to use L'hospital's rule but i dont know how to start this particular question.
I tried making the substitution y=1/x^2
but didnt get anywhere.
Any help or hints would be v much appreciated.
Thank you

2. Jan 4, 2009

### jgens

My only suggestion is is to use logarithmic differentiation and L'hospital's rule to solve the problem.

3. Jan 4, 2009

### sara_87

yes that sounds good.
but what do you mean by logarithmic differentiation??

Thanks

4. Jan 4, 2009

### jgens

Let y = (sin(x)/x)^(1/x^2), than log(y) = log(sin(x)/x)/(x^2). Since we may think of y as y = e^f(x), we need only find the limit as f(x) tends to zero. Does that make sense? As a side note, log designates the natural logarithm or logarithm base e.

5. Jan 4, 2009

### sara_87

so f(x)= log(sin(x)/x)/(x^2)

we need to find this limit??
doesnt this make things more diificult? i dont know mayb i dont get it. :s

6. Jan 4, 2009

### jgens

If you're required to use L'hospital's rule I think you may need to find that limit. You may need to apply the rule several times.

7. Jan 4, 2009

### sara_87

but, log(sin(x)/x)/(x^2) is not a fraction so how do we apply L'Hospitals?
What would be the first step

8. Jan 4, 2009

### NoMoreExams

Assuming the problem is to find

$$y = \lim_{x \rightarrow 0} \left( \frac{sin(x)}{x} \right)^{\frac{1}{x^2}}$$

Then, we can take ln(.) of both sides to get

$$ln|y| = \lim_{x \rightarrow 0} \frac{1}{x^2} ln \left( \frac{sin(x)}{x} \right)$$

Now we know that the limit inside the ln(.) tends to 1 and ln(1) = 0 so we effectively have a case of 0/0 and so we can use L'Hopital's.

Can you take it from here?

9. Jan 4, 2009

### jgens

It is a fraction: f(x) = log(sin(x)/x) and g(x) = x^2. Therefore f(x)/g(x) = log(sin(x)/x)/x^2.

10. Jan 4, 2009

### NoMoreExams

This is a pretty nasty limit but I'm pretty sure there's a trick. Plus knowing the answer makes me think the trick is to use Maclaurin if you know it and play around with derivative of sums.

Last edited: Jan 4, 2009
11. Jan 4, 2009

### sara_87

ok so if i do:
ln(y)=lim(1/x^2)ln(sinx/x)
and use L'hospital's,
i get:
ln(y)=lim (2xcosx-sinx)/sinx

using L'hospitals rule again, i get:

ln(y)= lim (cosx-2xsinx)/(cosx)

substitutin x=0 into the limit, i get:

ln(y) = 1

?? did i make a mistake somehere?

12. Jan 4, 2009

### NoMoreExams

Yes. What's the derivative of

ln(sin(x)/x)?

13. Jan 4, 2009

### jgens

Correct me if I'm wrong but shouldn't the first application of L'hosptial's rule yield: (xcos(x) - sin(x))/(2x^2sin(x))?

14. Jan 4, 2009

### sara_87

after simplifying it will be:
(xcosx - sinx)/(xsinx)

Oh sorry i made a mistake,
after applying the first L'hospitals, i should get:
ln(y)=lim (2xcosx-2sinx)/sinx

so applying again, i should get:
ln(y)=lim (-2xsinx)/cosx

right??
so do i have to apply L'hospitals rule again?

15. Jan 4, 2009

### jgens

I still don't agree with the answer on your first application; however, assuming you've done the steps correctly, you can no longer apply L'hospital's rule at the step you've arrived at.

d(ln(sin(x)/x))/dx = (x/sin(x))((xcos(x)-sin(x))/x^2) correct?

16. Jan 4, 2009

### NoMoreExams

That is correct. Why can't you apply L'Hopital's here? When you "plug in" x = 0, you get 0/0 again.

17. Jan 4, 2009

### jgens

His or her final step: -2xsin(x)/cosx is not in the form 0/0, that's what I was pointing to - he or she asked if another application of the rule was needed. Sorry for any confusion. Moreover, While he or she could substitute 0 into this step and obtain an answer, the result is not correct.

18. Jan 4, 2009

### sara_87

the first time i applied L'hospital's, i differentiate ln(sinx / x) divided by derivative of x^2

so: = (x/sinx)((xcosx-sinx)/x^2) cancel out an x to get: (1/sinx)((xcosx-sinx)/x)
divide this by 2x to give:.........oh ur right!!!!! after the first time, you would get:
ln(y)= (xcosx-sinx)/(2x^2sinx) ...like you said.
then after the second time, you would get:

ln(y)=(-sinx)/(4sinx+2xcosx)

right??

19. Jan 4, 2009

### sara_87

oh then after another time:
lim(-cosx)/(4cosx+2cosx-2xsinx)
so
ln(y)=-1/6

right?

20. Jan 4, 2009

### NoMoreExams

Yes that's the last step, and now since you are trying to find y... how do you get y = ?