Limit as $x\to 0$ of $x\sin(1/x)$
By Sarah Rodriguez •
How to find limit as $x \to 0$ of $x\sin(1/x)$?
For $x^2\sin(1/x)$, I know it's $0$ since by the Squeeze theorem, $-x^2 \le x^2\sin(1/x) \le x^2$, but for $x\sin(1/x)$, I run into some problems when applying Squeeze theorem.
$\endgroup$ 112 Answers
$\begingroup$To use the Squeeze Theorem, we do know that $0\leq|x \sin(1/x)|\leq|x|,\;$ so by the squeeze theorem $$\;\lim_{x\to 0} |x \sin(1/x)|=0,\;\implies \; \lim_{x\to 0} x \sin(1/x)=0.$$
$\endgroup$ 1 $\begingroup$Hint
For $x>0$ use the squeeze theorem and the inequality $$-x\leq x\sin\left(\frac{1}{x}\right)\leq x$$
Do the same thing for $x<0$ or better use the fact that the function $x\sin\left(\frac{1}{x}\right)$ is even.
