Squeeze Theorem to find $\lim_{x\to \infty}$ $$\lim_{x\to \infty} x^2 \sin\dfrac{1}{x^2}$$
The answer is 1 but I don't understand why. Can someone show me the steps they took to arrive at this solution?
 A: Hint:
$$
x^2\,\sin\frac1{x^2}=\dfrac{\sin\dfrac1{x^2}}{\dfrac1{x^2}}.
$$
A: Hint: Rewrite this expression as $$\frac{\sin\frac{1}{x^2}}{\frac{1}{x^2}}$$
and note that $\frac{1}{x^2}\rightarrow 0$ as $x\rightarrow\infty$.
A: If you want to squeeze,
use
$z-\frac{z^3}{3}
< \sin z 
< z
$
for
$0 < z < \pi/2$
(actually,
for all $z$
since it is an
enveloping series).
Then,
if 
$\frac1{x^2}
< \pi/2$,
or
$x > \sqrt{\frac{2}{\pi}}
$,
$\frac1{x^2} 
>\sin{\frac1{x^2}}
> \frac1{x^2}-\frac1{3x^4}
$
so
$1 
>x^2\sin{\frac1{x^2}}
> 1-\frac1{3x^2}
$.
To show
$z-\frac{z^3}{3}
< \sin z 
< z
$
for
$0 < z < \pi/2$,
start with
$\cos z < 1$
(which follows from
$\sin^2 z + \cos^2 z = 1$).
Integrate 
(all integrations are from
$0$ to $z$)
to get
$z > \sin z$.
Integrate again to get
$\frac{z^2}{2} > 1-\cos z$
or
$\cos z > 1-\frac{z^2}{2}$.
Integrate again to get
$\sin z > z-\frac{z^3}{3}
$.
(If you are worried about the
strict inequalities,
use non-strict - the result follows just as well.)
A: Another way to show this is the usage of l'hospital's rule:
$\lim_{x \to \infty } x^2 sin(\frac{1}{x^2}) = \lim_{x \to \infty } \frac{sin(\frac{1}{x^2})}{\frac{1}{x^2}} =\lim_{x \to \infty } cos(\frac{1}{x^2})*\frac{-2}{x^3}*\frac{x^3}{-2}=\lim_{x \to \infty } cos(\frac{1}{x^2})=cos(0)=1$
