How to use $\varepsilon$-$\delta$ in this equation? 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.
 A: $$|x^4\sin^2(1/x)|<|x^4|=x^4<\varepsilon$$
A: 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.
A: 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 $\ x \lt \epsilon^{(\frac {1}{4})} = \delta$ 
