Prove $\lim_{x \to0}{x\sin\left(\frac1{x}\right)} = 0$ by using $\epsilon$-$\delta$ method.

Question

I have done many exercises of proving limits by using this method, but this one, I don't know how can I prove it.

The definition says:

Prove $\displaystyle\lim_{x \to0}{x\sin\left(\displaystyle\frac{1}{x}\right)} = 0$ by using $\epsilon$-$\delta$ method.

Thank you very much.

For any given $\epsilon>0$ take $\delta=\epsilon$, then if $|x|<\delta$ you have that $|x\sin(1/x)|\leq |x|<\delta=\epsilon$ — conditionalMethod, Dec 04 '19 at 16:45

score 1 · Answer 1 · answered Dec 04 '19 at 16:47

1

Let $\epsilon > 0$. Notice that $|\sin\frac1x|\leqslant 1$ for all $x$, so for $|x-0|<\delta$, we have $$|x\sin\tfrac1x-0|=|x||\sin\tfrac1x|<\delta\cdot1=\delta,$$ so if we take $\delta=\epsilon$, we are done.

answered Dec 04 '19 at 16:47

Luke Collins

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Prove $\lim_{x \to0}{x\sin\left(\frac1{x}\right)} = 0$ by using $\epsilon$-$\delta$ method.

1 Answers1