# Limits and sine

1. Nov 5, 2007

### 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?

2. Nov 5, 2007

### d_leet

What is sin(0)?

3. Nov 5, 2007

### HallsofIvy

Staff Emeritus
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/$\sqrt{x}$?

4. Nov 5, 2007

### d_leet

I never claimed it would, but reading the original post the OP seems to have the idea that sin(0)=1.

5. Nov 6, 2007

### HallsofIvy

Staff Emeritus
Sorry, I see your point now!

6. Nov 6, 2007

### 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?).

7. Nov 6, 2007

### neutrino

L'Hopital's is not required here. A simple rearrangement of terms and proper use of limits will do.

Last edited: Nov 6, 2007
8. Nov 6, 2007

### Ignea_unda

Please show, at least how to set it up, because I am not seeing the rearrangement of the terms.

9. Nov 7, 2007

### HallsofIvy

Staff Emeritus
Then go back and read my first response!

10. Nov 7, 2007

### Math Jeans

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. Nov 7, 2007

### Ignea_unda

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

12. Nov 7, 2007

### d_leet

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?

13. Nov 8, 2007

### Ignea_unda

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

14. Nov 8, 2007