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Let us consider $f : \mathbb{R} \to \mathbb{R}$ defined by $$f(x) =\begin{cases} x^2 \sin \frac{1}{x},& x \neq 0\\ 0,& x = 0\end{cases}.$$

By using algebra of continuous functions (or algebra of limits) show that $f$ is continuous in $\mathbb{R}$. (You may assume without proving that $\sin y$ and polynomials are continuous at every point).

Okay so I understand that $x^2$ and $\sin\frac{1}{x}$ are both continuous functions, and due to algebra of continuous functions proving $f(x)$ is continuous but I'm not sure how to prove this? Thanks.

3SAT
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Lauren Bathers
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1 Answers1

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For $y\ne0$, $x\mapsto x^2$ and $x\mapsto \sin\frac1x$ are continuous at $y$, so their product is as well.

For $y = 0$, recall that $\left|\sin\frac1y\right|\leqslant 1$, so $|f(y)|\leqslant y^2$. Given $\varepsilon>0$, $|y|<\varepsilon^{\frac12}$ implies that $$|f(y)|\leqslant y^2<\varepsilon,$$ so that $f$ is continuous at $0$.

Math1000
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  • could you explain the y=0 bit to me. where did the $y^2$ come from, and why does proving it to be less than epsilon prove it continuous at 0? – Lauren Bathers Aug 12 '15 at 20:59
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    @LaurenBathers Do you remember the $\epsilon, \delta$ definition of continuity? If so, here the $\delta$ is $\sqrt{\epsilon}$. When $|y - 0| = |y| < \sqrt{\epsilon}$, then $|f(y) - f(0)| = |f(y) - 0| = |f(y)| < y^{2} < \sqrt{\epsilon}^{2} = \epsilon$. – layman Aug 12 '15 at 21:07
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    Also, since $\sin(\text{anything})$ is always $\leq 1$, then $x^{2} \sin(\frac{1}{x}) \leq x^{2}(1) = x^{2}$, right? That's why $|f(y)| \leq y^{2}$. As @Battani below me explains, this answer really only shows continuity at $x = 0$. But when $x \neq 0$, since $x^{2}$ is continuous at any $x$ and $\sin(\frac{1}{x})$ is continuous at any $x \neq 0$, then their product is also continuous at any $x \neq 0$. Here, we are using that the composition of continuous functions ($\sin(x)$ and $\frac{1}{x}$) and product of cts fcns ($x^{2}$ and $\sin(\frac{1}{x}$) are cts). – layman Aug 12 '15 at 21:08
  • what about contionus at R?OP asked to show it at R – haqnatural Aug 12 '15 at 21:08
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    @Battani Check out my comment above yours. – layman Aug 12 '15 at 21:11