How Do You Prove the Limit of sin(t)/sqrt(t) as t Approaches Infinity?

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Homework Statement



\lim_{t→∞}\frac{sin (t)}{\sqrt{t}}

Homework Equations


The Attempt at a Solution


This was actually part of a larger problem about improper integrals. The problem has been reduced to this, but I have no idea how to proceed from here. I know that sin(x) behaves very bizarrely at infinity, so I don't know if L'Hopital's rule can even be applied here.

My intuition tells me that the answer is 0, but how can we prove this? Must we refer to the ε-δ definition?

BiP
 
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Bipolarity said:

Homework Statement



\lim_{t→∞}\frac{sin (t)}{\sqrt{t}}

Homework Equations



The Attempt at a Solution


This was actually part of a larger problem about improper integrals. The problem has been reduced to this, but I have no idea how to proceed from here. I know that sin(x) behaves very bizarrely at infinity, so I don't know if L'Hopital's rule can even be applied here.

My intuition tells me that the answer is 0, but how can we prove this? Must we refer to the ε-δ definition?

BiP
Use the squeeze theorem.

What's \lim_{t→∞}\ 1/\sqrt{t}\ ?

How about giving us the entire problem?
 
Are you familiar with the squeeze theorem?
 
Ah, good old squeeze theorem why didn't I think of that?

Thanks guys!

BiP
 
The original problem (for Sammy):

\int^{π}_{0}\frac{dt}{\sqrt{t}+sin(t)}

BiP
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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