Answer to Question #229648 in Calculus for Live

Question

lim

x→0

x

2

sin 

Answer 1

$$\lim\limits_{x\to0}x^2\sin(\dfrac{1}{x})$$

$$-1\leq\sin(\dfrac{1}{x})\leq1$$

$$-x^2\leq x^2\sin(\dfrac{1}{x})\leq x^2$$

$$\lim\limits_{x\to0}x^2=\lim\limits_{x\to0}(-x^2)=0$$

Then by Squeeze Theorem

$$\lim\limits_{x\to0}x^2\sin(\dfrac{1}{x})=0$$