What is \(\displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right)\) equal to?

Concept:

Sandwich theorem: 

Let f(x), g(x) and h(x) be three real numbers having a common domain D such that g(x) ≤ f(x) ≤ h(x) ∀ x in some interval containing the point c. Then,

If \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{h}}\left( {\rm{x}} \right) = {\rm{L\;then\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{L}}\)

Calculation:

To Find: Value of \(\displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right)\)

As we know, \(\rm -1 \leq \sin (\dfrac{1}{x}) \leq 1\)

Multiplying by x², we get

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

\(\Rightarrow \displaystyle\lim_{\rm x\rightarrow 0} \rm (-x^2) \leq \displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right) \leq \displaystyle\lim_{\rm x\rightarrow 0} \rm x^2\)

\(\Rightarrow 0 \leq \displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right) \leq 0\)

∴ \(\displaystyle\lim_{\rm x\rightarrow 0} \rm x^2 \ sin \left(\dfrac{1}{x}\right)\)= 0