Lim of sqrt(x) * sine(1/sqrt(x)): Solving for x --> infinity

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In summary, the limit of sqrt(x) * sine(1/sqrt(x)) as x --> infinity is an indeterminate form. However, by rearranging terms and properly using limits, it can be simplified to 1 * sqrt(infinity). This may seem like it approaches infinity, but by applying L'Hopital's rule or using u-substitution, it can be shown that the limit is actually 1. This may be confusing when considering the value of sin(0), but it is important to remember that the limit is not the same as the actual value of the function at a specific point.
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Starcrafty
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I am doing a question that goes like this Lim of sqrt(x) * sine(1/sqrt(x)) as x --> infinity = ? what i determined was as x --> infinity 1/sqrt(x) would approach zero there for sine of 1/sqrt(x) would approach 1 there fore 1 * sqrt(infinity) would be infinity. however the answer says it is 1. can sum1 explain this?
 
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  • #2
What is sin(0)?
 
  • #3
Starcrafty said:
I am doing a question that goes like this Lim of sqrt(x) * sine(1/sqrt(x)) as x --> infinity = ? what i determined was as x --> infinity 1/sqrt(x) would approach zero there for sine of 1/sqrt(x) would approach 1 there fore 1 * sqrt(infinity) would be infinity. however the answer says it is 1. can sum1 explain this?

d_leet said:
What is sin(0)?
That wouldn't explain why the limit is 1 rather than 0!

Starcrafty, you've probably already proved that sin(x)/x goes to 1 as x goes to 0. What happens if you let u= 1/[itex]\sqrt{x}[/itex]?
 
  • #4
HallsofIvy said:
That wouldn't explain why the limit is 1 rather than 0!

I never claimed it would, but reading the original post the OP seems to have the idea that sin(0)=1.
 
  • #5
Sorry, I see your point now!
 
  • #6
Yes, what you actually have is an indeterminate form. As they said above, as x -> infinity, 1/sqrt(x) goes to 0. The sin(0) is not 1, but zero. Therefore you should try applying L'Hoppital's rule (theorem?).
 
  • #7
Ignea_unda said:
Yes, what you actually have is an indeterminate form. As they said above, as x -> infinity, 1/sqrt(x) goes to 0. The sin(0) is not 1, but zero. Therefore you should try applying L'Hoppital's rule (theorem?).
L'Hopital's is not required here. A simple rearrangement of terms and proper use of limits will do.
 
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  • #8
neutrino said:
L'Hopital's is not required here. A simple rearrangement of terms and proper use of limits will do.

Please show, at least how to set it up, because I am not seeing the rearrangement of the terms.
 
  • #9
Then go back and read my first response!
 
  • #10
neutrino said:
L'Hopital's is not required here. A simple rearrangement of terms and proper use of limits will do.

I agree that L'Hopital's rule is always easiest for limits that can support it. It is my main tool for all limits (excluding multivariable ones and limits that require other forms of solving and proof).
 
  • #11
HallsofIvy said:
That wouldn't explain why the limit is 1 rather than 0!

Starcrafty, you've probably already proved that sin(x)/x goes to 1 as x goes to 0. What happens if you let u= 1/[itex]\sqrt{x}[/itex]?

I'm sorry I'm being a pain, but what are you meaning "u" to be. Are you replacing 1/x with 1/[itex]\sqrt{x}[/itex]?
 
  • #12
Ignea_unda said:
I'm sorry I'm being a pain, but what are you meaning "u" to be. Are you replacing 1/x with 1/[itex]\sqrt{x}[/itex]?

He is meaning u to be 1/[itex]\sqrt{x}[/itex], what happens if you make this substitution? What does the limit become in terms of u?
 
  • #13
Hmmm, that's an interesting way to compute the limit. I've never seen it done with u-substitution before.
 
  • #14
Ignea_unda said:
I've never seen it done with u-substitution before.
You would have, if you had checked your PM Inbox. :wink:
 

1. What is the limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity?

The limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity is 0. This can be shown using the squeeze theorem and the fact that the limit of sine(1/sqrt(x)) as x approaches infinity is 0.

2. How do you solve for x when taking the limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity?

To solve for x, set the expression inside the sine function equal to zero. This will result in x = 0. Then, use the substitution method to show that the limit is equal to 0.

3. Can the limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity be evaluated using L'Hopital's rule?

No, L'Hopital's rule cannot be used for this limit because it is an indeterminate form (0 * infinity).

4. What is the significance of the limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity?

The limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity is significant in that it shows the behavior of the expression as x gets larger and larger. It also shows that the product of a function that approaches 0 and a function that approaches infinity can still have a finite limit.

5. How does the limit of sqrt(x) * sine(1/sqrt(x)) as x approaches infinity relate to real-world applications?

This limit can be seen in various areas of physics and engineering, such as in the study of wave propagation and oscillatory systems. It also has applications in the study of asymptotic behavior of functions and can be used to solve certain integrals and differential equations.

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