# show that the limit of $x \sin(1/x) = 0$ [duplicate]

Say we let $f$ be a real-valued function, let $S\subseteq \text{dom}(f)\subseteq R$, let $a\in \bar{S}$, and let $L\in R$.

We know the $\delta -\epsilon$ condition for $\lim_{x\to a} f(x)=L$ is:

$$\forall \epsilon >0: \exists \delta >0: \forall x\in S: |x-a|<\delta \to |f(x)-L|<\epsilon.$$

So given all that, how can we show $\lim_{x\to 0}x \cdot \sin(\frac 1 x )=0$ by proving the condition I wrote above, for $a=0$, $S=(0,\infty)$, $f(x)=x\cdot \sin(\frac 1 x)$ for all $x\in (0,\infty)$, and $L=0$?

• Use that the sine is bounded. Commented Nov 9, 2015 at 21:52

Given $\epsilon>0$, choose $\delta=\epsilon$. Then, for all $x$ with $|x|<\delta$, we have: $$\left|x\cdot\sin\left(\frac{1}{x}\right)\right|\leq|x|<\epsilon.$$
If $\lim_{x\to x_0}f(x)=0$, and $g(x)$ is bounded in some (possibly punctured) interval around $x_0$, then $\lim_{x\to x_0}f(x)g(x)=0$.
Let $\epsilon > 0$ be given. Notice that $$\Big|x\sin \frac{1}{x}\Big| \le |x| \cdot 1.$$ Then the desired inequality holds if $|x| < \epsilon =: \delta$.