This limit supposed to be $0$ but I get $2$, why? I am told that
$$
\lim_{x\to0}\left(4x^2\sin^2\left(\frac{1}{x}\right)-x\sin\left(\frac{2}{x}\right)\right)=0,
$$
but when I calculate this by hand, I get $2$, why? I thought that this limit is
$$
4\lim_{x\to0}\left(\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\right)^2-2\lim_{x\to0}\frac{\sin\frac{2}{x}}{\frac{2}{x}}=4-2=2.
$$
What I am doing wrong? And in general how can I notice that the method I used is not correct?
 A: You'll need to use the squeeze theorem (sandwich theorem) on each term here:
$-4x^2 \le 4x^2\sin^2\left(\frac{1}{x}\right) \le 4x^2$
and 
$-x \le x\sin\left(\frac{1}{x}\right) \le x$
What can you say about $ \displaystyle\lim_{x\to 0} \pm 4x^2$ and $\displaystyle\lim_{x \to 0} \pm x$ ?
A: $\lim x$ tends to $0$, $\dfrac{\sin x}{x}=1$
not $(\sin\dfrac1x)/\dfrac1x$ which is $=\ 0$.
A: Because : $\dfrac{\sin\infty}{\infty}=0\ne\infty$ because denominator is very large.
Obviously all those values are tending and not exact.
A: Another way is use $\lim f.g=\lim f\cdot \lim g \tag{1}$.
This is only true if the individual limits, $\lim f$ and $\lim g$ are finite.
 In our case $f(x)=x^2$ and $g(x)=\sin\left(\dfrac1{x^2}\right)$.
We wish to find $\lim\limits_{x\to 0} f(x)\cdot  g(x)$. We need to examine the individual limits here. As  $x^2$ tends to $0$ as $x\to 0$ so $\lim f(x)$ exists. Now the question is does $\lim\limits_{x\to 0}\sin\left (\dfrac1x \right)$ exists. No, it doesn't but it is finite because $\sin$ function always lies in $[-1,1]$. So
$$\lim_{x\to 0} f(x)\cdot g(x)=\lim_{x\to 0} f(x) \lim_{x\to 0} g(x)$$
$$\lim_{x\to 0} x^2\cdot \sin\left(\dfrac1{x^2}\right)=\lim_{x\to 0} x^2\cdot \lim_{x\to 0}\sin\left(\dfrac1{x^2}\right)=0\cdot \text{a finite quantity}$$
As we know $0$ into a finite value is $0$, so the limit of $f(x)g(x)$ as $x\to0$ is $0$.
