# Use limit theorems to prove $\sqrt{x}\sin\left(\frac{1}{x}\right)$ is continuous on $[0, 1]$

Use limit theorems to prove continuous on $$[0,1]$$

$$f(x) = \begin{cases} \sqrt{x}\sin(\frac{1}{x}), & \text{if x \neq 0} \\ 0, & \text{if x=0} \end{cases}$$

The only problem point is $$0$$ clearly $$\sin(\frac{1}{x})$$ is discontinuous at $$0$$. when $$x=0, f(x)=0$$

I'm a little confused my notes showed a similar problem but that problem was factorable and plugged in the maximum value and got the same value in the interval if $$f(x)=0$$.

I am not really sure what you mean by "factorabale", but I guess your example went something like

$$|\sin (1/x)|\leq 1 \quad \text{for all } x\neq 0$$

Then for $$x\neq 0$$ $$|\sqrt x\sin (1/x)|\leq \sqrt x$$ and as $$x\to 0$$ we have $$\sqrt x\to 0$$. Then by the squeeze theorem you get $$\sqrt x\sin (1/x)\to 0$$

Let $$\epsilon >0$$ be given, and $$0;

$$|√x\sin (1/x)|\le √x$$ ;

Choose $$\delta = \epsilon^2$$.

Then

$$0 implies

$$|√x\sin(1/x)| \le √x < √\delta =\epsilon$$, i.e.

$$\lim_{x \rightarrow 0^+}f(x)=f(0)$$.