Evaluating infinite limits

[CaptainOfSmug]

Hello all, first time posting!

So my question I have been presented with is:

limit as x approaches infinity sqrt(x)*sin(1/x)

I'm not really sure how to go about this problem, I want to say the answer is 0 but can't prove it... so obviously I'm not getting the concept here. I have a feeling I should maybe use squeeze theorem but am not sure exactly how that would help me. Anyways, if someone could explain this problem to me it would be greatly appreciated! I've been stuck on it for about an hour now....

 

[Quaid]

limit as x approaches infinity sqrt(x)*sin(1/x)

I want to say the answer is 0 but can't prove it...

If you're thinking that the product is zero because sin(1/x) is headed toward zero, then you need to remember that sin(1/x) will never be zero. It will be a very small number, but the other factor sqrt(x) is heading toward infinity. So, in these type of tug-of-war scenarios between opposing numbers, we need to be careful not to jump to conclusions. It comes down to whether sqrt(x) is growing faster than sin(1/x) is shrinking.

so obviously I'm not getting the concept here. I have a feeling I should maybe use squeeze theorem but am not sure exactly how that would help me.

There is no single concept to get here. You need to think about different limit methods that you've learned. You considered The Squeeze Theorem, and that shows effort.

If you were able to come up with two functions which both have a limit of zero, as x approaches infinity, AND the value of sqrt(x)*sin(1/x) is always between those two functions, then your limit must also be zero. (It's often not easy to find such functions, though.)

Another method is L'Hôpital's Rule. Have you seen that one, yet?

The given function can be rewritten as a ratio with sin(1/x) in the numerator. Can you see what the denominator would be?

This ratio approaches 0/0, as x approaches infinity, and so it's a candidate for L'Hôpital's Rule.

Let us know.