Why are the two results of this limit inconsistent?

Question

It is a well know result that $$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$ so we should have the result that $$\lim_{x\rightarrow 0} x^2\sin \left(\frac{1}{x^2}\right)=1$$ yet when I plug the limit into an online calculator it uses the squeeze theorem and shows that $$-x^2 \leq x^2\sin \left(\frac{1}{x^2}\right) \leq x^2 $$ because of the values $\sin$ can take. This leads us to the result that the limit is zero because of the limits of $x^2,-x^2$ as $x$ tends to zero.

Why is the middle line wrong though? I've been sat for a while thinking about it but I can't see why I'm guessing there is something obvious I'm not spotting.

Thanks.

Answer 1

Robert has the correct answer in the comments: $x\to 0$ in the first limit corresponds to $x\to \infty$ in the second limit, so the analogy is incorrect.

Answer 2

$$\lim_{x\rightarrow 0} x^2\sin \left(\frac{1}{x^2}\right)=0$$ because $\left|\sin \left( \frac{1}{x^2} \right) \right|\leq 1$ and $\lim\limits_{x\rightarrow 0} x^2=0$

Regarding $\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}=1$ For one that knows the L'Hospital Rule it is simple $$\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}= \lim\limits_{x\rightarrow 0} \frac{\cos(x)}{1}=1$$

Without the L'Hospital Rule go to Link it user some geometry basics and squeeze theorem, shouldn't not that difficult to understand. Another option is Another Link.

Answer 3

As requested, I'm making my comment into an answer:

$x \to 0$ in the first limit corresponds to $x \to \infty$ in the second.

I should make that a bit more precise. It is better to use a different name for the variable in the other limit. With $t = 1/\sqrt{|x|}$ $$\eqalign{\lim_{x \to 0+} \dfrac{\sin(x)}{x} &= \lim_{t \to \infty} t^2 \sin(1/t^2)\cr \lim_{x \to 0-} \dfrac{\sin(x)}{x} &= \lim_{t \to \infty} -t^2 \sin(-1/t^2) = \lim_{t \to \infty} t^2 \sin(1/t^2) \cr}$$