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Let the function $f(x)$ be defined as: $$ f(x) = \begin{cases} \sin {\frac {1}{x}},& \text{if } x \neq 0 \\ 0 , & \text{if } x= 0 \end{cases} $$

Is this function continuous at $ x=0$ ?

Lakhi
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  • In every neighborhood of $0$ $\sin\frac1x$ is both positive and negative, so $f$ is not continuous at $0$. – rubik Feb 10 '15 at 11:06
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    Actually, the condition of being positive and negative in every neighborhood of $0$ does not preclude continuity at $0$. (However, in this case, I agree that the function is not continuous at $0$). – paw88789 Feb 10 '15 at 11:14

2 Answers2

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No. If you let $x_n=\frac1{n\pi}$ for each $n\ge1$, then $f(x_n)=0$, but if you let $y_n=\frac2{(2n+1)\pi}$ then $f(y_n)=1$. Since both $x_n,y_n\to0$ as $n\to\infty$, $f(x)$ does not approach a single limit as $x\to0$. In particular it is not continuous at $x=0$.

Jason
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Hint. Consider the sequences $$ x_n=\frac{1}{2n\pi+\pi/2}\quad\text{and}\quad y_n=\frac{1}{2n\pi-\pi/2}. $$ Then $$ x_n,y_n\to 0, $$ but $$ f(x_n)=1\quad\text{while}\quad f(y_n)=-1. $$

Yiorgos S. Smyrlis
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