Find limit $x\rightarrow0$ of $f(x)=x^2\cdot\left({\sin{\frac 1 x}}\right)^2$

Question

I have following function:

$$f(x)=x^2\cdot\left({\sin{\frac 1 x}}\right)^2$$

I want to find the limit of the function for $x\rightarrow0^\pm$. First I analyze $\frac 1 x$:

• $\frac {1}{x}\rightarrow +\infty$ for $x\rightarrow0^+$

but the $\sin$ of infinity does not exist. Then I use the comparison theorem (I don't know how it's called in English) and conclude that, because

$$\left|{x^2\left({\sin{\frac 1 x}}\right)}^2 \right| \le \frac{1}{x^2}\rightarrow0^+$$

therefore the initial function tends to $0$. Is this reasoning correct? Are there better ways?

Answer 1

If you meant $\left\lvert x^2\sin^2\left(\frac1x\right)\right\rvert\leqslant x^2$, then yes, it is correct. It follows from this that $\lim_{x\to 0}\left\lvert x^2\sin^2\left(\frac1x\right)\right\rvert=0$ and that therefore $\lim_{x\to 0}x^2\sin^2\left(\frac1x\right)=0$.

Answer 2

Your argument is correct and the result is OK.

Of course you meant $$\left|{x^2\left({\sin{\frac 1 x}}\right)}^2 \right| \le x^2\rightarrow0$$

Please edit your question accordingly.