## University Calculus: Early Transcendentals (3rd Edition)

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$
$$\lim_{x\to0}x^2\sin\frac{1}{x}=0$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$ 1) We know that for all $x$, $$|\sin\frac{1}{x}|\le1$$ Therefore, $$|x^2\sin\frac{1}{x}|\le|x^2|$$ 2) Now let an arbitrary value of $\epsilon\gt0$, and let a value of $\delta$ so that $\delta=\min(1,\epsilon)$. There are only 2 cases: - If $\epsilon\le1$, then $\delta=\epsilon\le1$, so $\delta^2\le\delta=\epsilon$. - If $\epsilon\gt1$, then $\delta=1\lt\epsilon$, so $\delta^2=1\lt\epsilon$ Therefore, $\delta^2\le\epsilon$ For all $x$, as $0\lt|x|\lt\delta$, we have $$|f(x)|=|x^2\sin\frac{1}{x}|\le|x^2|\lt\delta^2$$ $$|f(x)|\lt\delta^2$$ Yet since $\delta^2\le\epsilon$, $$|f(x)|\lt\epsilon$$ The proof has been completed.