Limit at Infinity: Solve Without Squeeze Theorem

[mateomy]

Homework Statement

<br /> \lim_{x \to \infty} \sqrt{x}\sin\frac{1}{x}<br />

Homework Equations

I don't think you can use the squeeze theorem here...

The Attempt at a Solution

So I am just studying for an exam that I have tomorrow and I am going through problems that weren't assigned on our homework set, (just in case he wants to slip something in there).

I was looking at the solution to the fore-mentioned limit equation and it references the expansion of sinx. Is this necessary? I've used substitution before in problems like lim as x approaches infinity of xsin(1\x) setting 1\x as y while t approaches zero. Can that sort of thing be done here?THANKS FOR ANY HELP!

 

[kof9595995]

To make things clearer:
let t = \frac{1}{x}
an now the limit is
\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{{\sqrt t }} = \mathop {\lim (}\limits_{t \to 0} \frac{{\sin t}}{t} \times \sqrt t ).
Now you shall see how to do it.

 

[mateomy]

Thank you!