# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 49

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$

#### Work Step by Step

$$\lim_{x\to0}x\sin\frac{1}{x}=0$$ 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, $$|x\sin\frac{1}{x}|\le|x|$$ 2) Now let an arbitrary value of $\epsilon\gt0$, and let a value of $\delta$ so that $\delta=\epsilon$. For all $x$, as $0\lt|x|\lt\delta$, we have $$|f(x)|=|x\sin\frac{1}{x}|\le|x|\lt\delta$$ $$|f(x)|\lt\delta$$ Yet since $\delta=\epsilon$, $$|f(x)|\lt\epsilon$$ The proof has been completed.

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