Proving continuity using $\delta $- $\varepsilon$

Question

The question is:

Given the function $f:[-1,1]\to\Bbb R$, defined by $$f(x)= \begin{cases}x\sin\frac{1}{x},& \text{for }x\neq 0\\ 0,&\text{for }x=0. \end{cases}$$

Prove that the function admits maximum and minimum in its domain (Note that you are not required to compute the maximum and minimum only to prove their existence).

I've started off using $$|f(x) - f(x_0)| < \varepsilon {\ } \forall x\in[-1,1]{\ } \text{ and }{\ }|x-x_0|<\delta$$

This gives me $$|f(x)-f(x_0)| = \left|x\sin\left(\frac{1}{x}\right) - 0\right| = \left|x\sin\left(\frac{1}{x}\right)\right| < \varepsilon{\ } \text{ as } {\ }f(x_0)=f(0)=0$$

However I don't know where to go from here so any suggestions would be really useful. Thank you!

Answer 1

Let $\epsilon > 0$, $x_0 := 0$. Set $\delta := \epsilon$ and let $x \in (-\delta, \delta) \setminus \{0\}$. Then

$$|f(x) - f(x_0)| = \left|x \cdot\sin\left(\frac{1}{x}\right)\right| = \underbrace{|x|}_{< \delta}\cdot \underbrace{\left|\sin\left(\frac{1}{x}\right)\right|}_{\leq 1} < \delta = \epsilon$$

Answer 2

Use $|\sin x| \leq 1$. You can also simply use the squeeze theorem if you don't want to resort to epsilons and deltas.

Answer 3

You've proved that $f$ is continuous on the compact interval $[-1,1]$. Hence the image of that interval under $f$ is a compact interval itself.