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?

  • 2
    $\begingroup$ Did you want to write $\left|{\left(x^2\sin^2{\frac 1 x} \right)}\right|\le {x^2}\rightarrow0$? $\endgroup$ – Kusma Aug 21 '18 at 14:21
  • $\begingroup$ Yes exactly @Kusma thanks $\endgroup$ – Cesare Aug 21 '18 at 14:23
  • $\begingroup$ The function in the title is not quite the same as that in the question and You should correct line 7 as suggested by kusma. You reasoning is correct. $\endgroup$ – Peter Melech Aug 21 '18 at 14:28
  • $\begingroup$ Hint: Use L'hopital's rule.. $\endgroup$ – ChinG Aug 21 '18 at 14:29
  • 1
    $\begingroup$ you can also use: $0\le x^2\sin^2 \frac{1}{x}\le x^2\to 0$ as $x\to 0$. $\endgroup$ – farruhota Aug 21 '18 at 14:50

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 $\lim_{x\to 0}x^2\sin^2\left(\frac1x\right)=0$.


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.


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