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

[Starcrafty]

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]

What is sin(0)?

 

[HallsofIvy]

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?

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/\sqrt{x}?

 

[d_leet]

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.

 

[HallsofIvy]

Sorry, I see your point now!

 

[Ignea_unda]

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?).

 

[neutrino]

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.

 

[Ignea_unda]

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.

 

[HallsofIvy]

Then go back and read my first response!

 

[Math Jeans]

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).

 

[Ignea_unda]

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/\sqrt{x}?

I'm sorry I'm being a pain, but what are you meaning "u" to be. Are you replacing 1/x with 1/\sqrt{x}?

 

[d_leet]

I'm sorry I'm being a pain, but what are you meaning "u" to be. Are you replacing 1/x with 1/\sqrt{x}?

He is meaning u to be 1/\sqrt{x}, what happens if you make this substitution? What does the limit become in terms of u?

 

[Ignea_unda]

Hmmm, that's an interesting way to compute the limit. I've never seen it done with u-substitution before.

 

[neutrino]

I've never seen it done with u-substitution before.

You would have, if you had checked your PM Inbox.