Proof, EXT2 EQ-Bank 21 Prove that \(\lim\limits_{x \to 0}x^2\, \sin \left(\dfrac{1}{x}\right)=0\). (2 marks) --- 5 WORK AREA LINES (style=lined) --- Show Answers Only \(-1 \leqslant \sin \left(\dfrac{1}{x}\right) \leqslant 1\) \(-x^2 \leqslant x^2\, \sin \left(\dfrac{1}{x}\right) \leqslant x^2\) \(\text{By squeeze theorem:}\) \(\text{Since} \ \ \lim\limits_{x \to 0}-x^2=0\ \ \text{and}\ \ \lim\limits _{x \rightarrow 0} x^2=0\) \(\Rightarrow \lim\limits _{x \rightarrow 0} x^2\, \sin \left(\dfrac{1}{x}\right)=0\) Show Worked Solution \(-1 \leqslant \sin \left(\dfrac{1}{x}\right) \leqslant 1\) \(-x^2 \leqslant x^2\, \sin \left(\dfrac{1}{x}\right) \leqslant x^2\) \(\text{By squeeze theorem:}\) \(\text{Since} \ \ \lim\limits_{x \to 0}-x^2=0\ \ \text{and}\ \ \lim\limits _{x \rightarrow 0} x^2=0\) \(\Rightarrow \lim\limits _{x \rightarrow 0} x^2\, \sin \left(\dfrac{1}{x}\right)=0\)