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First Principle of Differentiation

Evaluate li...

Question

Evaluate $lim x \to 0 x^{2} sin \frac{1}{x}$

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Solution

$\begin{matrix} W e k n o w f r o m t r i g o n o m e t r y t h a t - 1 \leq sin (\frac{1}{x}) < - 1 f o r a l l x \neq 0 f o r {lim}_{x \to 0} w e d o n^{'} t c a r e w h a t h a p p e n s w h e n x = 0 s i n c e x < 2 > 0 f o r a l l x \neq 0, w e c a n m u l t i p l y t h r o u g h b y x^{2} t o g e t - x^{2} = x^{2} sin (\frac{1}{x}) \leq x^{2} C l e a l y {lim}_{x \to 0} (- x^{2}) = 0 a n d {lim}_{x \to 0} x^{2} = 0, s o, b y t h e s q u e e z e t h e o r e m . {lim}_{x \to 0} x^{2} sin (\frac{1}{x}) = 0 H e n c e, t h e r e q u i r e d a n s w e r i s 0. \end{matrix}$

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