What is the Limit of (sinx/x)^(1/x^2) as x Approaches 0 Using L'Hospital's Rule?

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In summary, to calculate the limit as x tends to 0 of (sinx/x)^(1/x^2), use logarithmic differentiation and L'hospital's rule to simplify the expression to ln(y) = 1. Then, apply L'hospital's rule twice more to get ln(y) = -1/6. Finally, solve for y to get y = e^(-1/6).
  • #1
sara_87
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Homework Statement



Calculate the limit as x tends to 0:

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

Homework Equations





The Attempt at a Solution



I have to use L'hospital's rule but i don't 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
 
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  • #2
My only suggestion is is to use logarithmic differentiation and L'hospital's rule to solve the problem.
 
  • #3
yes that sounds good.
but what do you mean by logarithmic differentiation??

Thanks
 
  • #4
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
so f(x)= log(sin(x)/x)/(x^2)

we need to find this limit??
doesnt this make things more diificult? i don't know mayb i don't get it. :s
 
  • #6
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
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
sara_87 said:

Homework Statement



Calculate the limit as x tends to 0:

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

Homework Equations


The Attempt at a Solution



I have to use L'hospital's rule but i don't 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

Assuming the problem is to find

[tex] y = \lim_{x \rightarrow 0} \left( \frac{sin(x)}{x} \right)^{\frac{1}{x^2}} [/tex]

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

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

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
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
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:
  • #11
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
sara_87 said:
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?

Yes. What's the derivative of

ln(sin(x)/x)?
 
  • #13
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
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
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
jgens said:
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))?

That is correct. Why can't you apply L'Hopital's here? When you "plug in" x = 0, you get 0/0 again.
 
  • #17
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
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
oh then after another time:
lim(-cosx)/(4cosx+2cosx-2xsinx)
so
ln(y)=-1/6

right?
 
  • #20
sara_87 said:
oh then after another time:
lim(-cosx)/(4cosx+2cosx-2xsinx)
so
ln(y)=-1/6

right?

Yes that's the last step, and now since you are trying to find y... how do you get y = ?
 
  • #21
Edit: Your result for ln(y) is correct. Now simply find the limit for y.
 
  • #22
so y=exp(-1/6)
yes?
 
  • #23
Yes :)
 
  • #24
Thank you v much
;)
 

FAQ: What is the Limit of (sinx/x)^(1/x^2) as x Approaches 0 Using L'Hospital's Rule?

What is L'hospital's rule?

L'Hospital's rule is a mathematical method used to evaluate limits of indeterminate forms, where both the numerator and denominator approach zero or infinity. It can be applied to a variety of functions, including rational, exponential, and trigonometric functions.

When should L'hospital's rule be used?

L'Hospital's rule should only be used when the limit of a function cannot be easily evaluated using other methods, such as direct substitution or factoring. It is typically used for more complex functions with indeterminate forms, and should not be used as a shortcut for every limit problem.

Are there any limitations to using L'hospital's rule?

Yes, there are limitations to using L'Hospital's rule. It can only be applied to indeterminate forms, where both the numerator and denominator approach zero or infinity. It also cannot be applied to limits at infinity, as the rule requires a finite limit in order to work.

Can L'hospital's rule be used for all types of functions?

No, L'Hospital's rule can only be applied to certain types of functions, such as rational, exponential, and trigonometric functions. It cannot be applied to functions with discontinuities, non-differentiable points, or infinite oscillations.

How do I know if I have correctly applied L'hospital's rule?

To ensure that L'Hospital's rule has been correctly applied, it is important to check the original function and the resulting limit. If the original function and the limit both approach the same value, then L'Hospital's rule has been applied correctly. Additionally, it is always helpful to double check the result using other methods, such as graphing or direct substitution.

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