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I'm trying to prove that

$$f\left(x\right)=\begin{cases} \frac{x^{2}}{1+\sin^{2}\left(\frac{1}{x}\right)}, & x\neq0\\ 0, & x=0 \end{cases}$$

is continuous at 0.

I know that I should show that $$\lim_{x\rightarrow0}\frac{x^{2}}{1+\sin^{2}\left(\frac{1}{x}\right)}=0$$

But I don't understand what is $$\lim_{x\rightarrow0}\sin^{2}\left(\frac{1}{x}\right)=?$$ When $x$ goes to $0$, $\dfrac{1}{x}$ goes to infinity, but sine oscillates between $-1$ and $1$, hence I don't know what limit would be.

Could you, please, help me?

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  • $\begingroup$ $\lim_{x\rightarrow 0}\sin\left(\frac{1}{x}\right)$ doesn't exist. This is known as the topologist's sine curve. $\endgroup$ – Cameron Williams Jun 27 '13 at 17:49
  • $\begingroup$ Note $\sin^{\color{maroon}2}(1/x)\ge 0$. $\endgroup$ – David Mitra Jun 27 '13 at 17:51
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You should apply squeeze theorem here. We have that for $x\in\mathbb{R}\setminus \{0\}$, $-1 \le \sin\left(\frac{1}{x}\right) \le 1$. So we have that $0 \le \sin^2\left(\frac{1}{x}\right) \le 1$. Let's make use of this property.

$$0 \le \lim_{x\rightarrow 0} \frac{x^2}{1+\sin^2\left(\frac{1}{x}\right)} \le \lim_{x\rightarrow 0} \frac{x^2}{1+0}.$$

Applying the limit on the right, we have that the overall limit is $0$.

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  • $\begingroup$ Amazing! I completely forgot about this theorem. Thank you, Cameron $\endgroup$ – Vovan Jun 27 '13 at 17:58
  • $\begingroup$ You're welcome. The squeeze theorem is very, very useful in limits involving trig functions. Often if it's not immediately clear how to do a certain limit with trig functions, squeeze theorem will be of use. $\endgroup$ – Cameron Williams Jun 27 '13 at 17:59
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Hint: Use the fact that $\frac{1}{1+\sin^2(1/x)}$ is always bounded, to prove continuity.

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