How to use $\varepsilon$-$\delta$ in this equation?

Question

How to use $\varepsilon$-$\delta$ in this equation?

$$\lim_{x \to 0}\;x^4\sin^2\left(\frac{1}{x}\right)$$

I can understand how to modify $\epsilon$ in this equation.

How should I handle the $\sin^2\left(\frac{1}{x}\right)$ in this question?

Thanks for your help. I wish for you to be safe from the corona virus.

Answer 1

$$|x^4\sin^2(1/x)|<|x^4|=x^4<\varepsilon$$

Answer 2

We have to prove that $$ \forall \varepsilon > 0\exists \delta \left( \varepsilon \right) > 0:0 < \left| x \right| < \delta \left( \varepsilon \right) \Rightarrow \left| {x^4 \sin ^2 \frac{1} {x}} \right| < \varepsilon $$ Since, for every $x\in \mathbb R\setminus \{0\}$ it is $ 0 \leqslant \sin ^2 \frac{1} {x} \leqslant 1 $ we consider the inequality $x^4 \leq \varepsilon$. We get $$ - \sqrt[4]{\varepsilon } \leqslant x \leqslant \sqrt[4]{\varepsilon } $$ Therefore , if $\delta(\varepsilon)=\sqrt[4]{\varepsilon }$ we have that $$ 0 \leqslant x^4 \sin ^2 \frac{1} {x} \leqslant x^4 \leqslant \varepsilon $$ for every $x \in (-\delta(\varepsilon),\delta(\varepsilon)) \setminus \{0\}$. By definition this proves what is required.

Answer 3

Let $\ f(x)$= $\ x^{4} sin^{2}(\frac {1}{x})$. Then $$\ |f(x)-0|=|x^{4} sin^{2}(\frac {1}{x})|$$ $$\leq |x^{4}|(since \ |sin^{2}(\frac {1}{x})| \leq 1)$$ $$\ = x^{4}$$ $$\lt \epsilon $$ whenever $\ x^{4} \lt \epsilon (=\delta^{4})$. Thus $$\ |f(x)-0| \le \epsilon $$ whenever $\ x \lt \epsilon^{(\frac {1}{4})} = \delta$