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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 $f(x)=0$.

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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 $\sqrt x\sin (1/x)\to 0$

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Let $\epsilon >0$ be given, and $0<x \le 1$;

$|√x\sin (1/x)|\le √x$ ;

Choose $ \delta = \epsilon^2$.

Then

$0<x<\delta $ implies

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

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

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