Exploring the Limit of $\frac{1}{x} \int_0^x (1-tan2t)^{\frac{1}{t}}dt$

  • Thread starter tangur
  • Start date
  • Tags
    Limit
In summary, the conversation involves a question on how to solve the problem \lim_{x \rightarrow 0} \frac{1}{x} \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt, with suggestions to use l'Hopital's rule and take the derivatives. The correct solution is to first take the integral, and then use the limit definition of e^x to solve for the limit, which is e^-2.
  • #1
tangur
14
0
[tex]\lim_{x \rightarrow 0} \frac{1}{x} \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt[/tex]

and nevermind the { in front of the 0, i couldn't figure out how to take it out, its my first time posting

I can't figure out how to start attacking this problem, do I have to intregrate by parts? if so what do I use as u and v.
I tried thinking of any identities that could help but nothing came to mind.

Any help will be greatly appreciated

Thanks in advance.

Evaluating it with maple gives [tex] e^{-2} [/tex]

Admin note: pesky { removed. Click the latex image to see its code.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Well, you could re-write it as:
[tex]\lim_{x\rightarrow 0} \frac{ \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt}{x}[/tex]
And then apply l'Hopital's rule.

No clue if that's the right way to go, and I'm too lazy to check.
 
  • #3
you would apply lhopital, take the limit and then diifferentiate?
 
  • #4
oh damn just as i was writing this message i figured out how to apply lhopital , so simple but it escaped by grasp, its just a differentiation of an integral, thanks so much
 
  • #5
[tex]\lim_{x\rightarrow 0} \frac{ \int_0^x {(1-tan2t)}^{\frac{1}{t}} dt}{x}[/tex]
Take the derivatives of the top and bottom:
[tex]\lim_{x \rightarrow 0} \frac{(1-tan(2x))^{\frac{1}{x}}}{1}[/tex]
[tex]\lim_{x \rightarrow 0} (1 - \tan (2x))^{\frac{1}{x}}[/tex]

But now, the expression inside the parens goes to 1 since [tex]\tan(0)=0[/tex] and the exponent should also tend to bring things to 1, so the limit should be 1.
 
  • #6
This is wrong, Nate G!
You've got the indeterminate form [tex]1^{\infty}[/tex]
we have:
[tex]\lim_{x\to0}(1-tan(2x))^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{ln(1-tan(2x))}{x}}[/tex]
[tex]=e^{\lim_{x\to0}\frac{-2x}{x}}=e^{-2}[/tex]
 
  • #7
arildno said:
This is wrong, Nate G!
You've got the indeterminate form [tex]1^{\infty}[/tex]
we have:
[tex]\lim_{x\to0}(1-tan(2x))^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{ln(1-tan(2x))}{x}}[/tex]
[tex]=e^{\lim_{x\to0}\frac{-2x}{x}}=e^{-2}[/tex]

Stupid brain...for some strange reason I thought it was going to [tex]1^0[/tex]. <Puts on pointy hat with donkey ears.>
 
  • #8
just take the integral first then it will be real easy to do this question
 

1. What is the purpose of exploring the limit of $\frac{1}{x} \int_0^x (1-tan2t)^{\frac{1}{t}}dt$?

The purpose of exploring this limit is to understand the behavior of the function as the independent variable, x, approaches infinity. This can provide insights into the characteristics and properties of the function and its relationship to other mathematical concepts.

2. What is the mathematical significance of this limit?

This limit is important in calculus because it involves the integration of a transcendental function, tan2t, and the reciprocal of the independent variable. It can also be used to evaluate other integrals and solve related problems in mathematics and physics.

3. How is this limit evaluated?

This limit can be evaluated using various techniques such as L'Hopital's rule, Taylor series expansion, or substitution of variables. The chosen method may depend on the complexity of the function and the desired level of accuracy in the result.

4. What are the potential applications of this limit?

This limit has applications in areas such as differential equations, Fourier analysis, and probability theory. It can also be used in the study of growth and decay models, optimization problems, and other mathematical and scientific fields.

5. How does the value of this limit change with different values of x?

The value of this limit may vary depending on the specific value of x. For example, as x approaches 0, the limit may tend to a specific value or may not exist at all. As x gets larger, the limit may approach infinity or converge to a finite value. Understanding these variations can provide insights into the behavior of the function and its relationship to other mathematical concepts.

Similar threads

Replies
1
Views
1K
Replies
29
Views
2K
Replies
2
Views
1K
Replies
5
Views
1K
Replies
3
Views
1K
Replies
33
Views
2K
Replies
4
Views
1K
Replies
4
Views
1K
  • Math POTW for University Students
Replies
3
Views
589
Back
Top