Question

Determine if \[f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}\] is a continuous function?

 

Answer 1

The given function f is \[f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}\]

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

` " If  c ≠ 0 , then "f(c)= c^2 sin (1/c)`

`lim_(x->c)f(x)lim_(x->c)(x^2 sin  1/x)=( lim_(x->c)x^2 )(lim_(x->c)sin 1/x)=c^2 sin (1/c)`

`∴lim_(x->c)f(x)=f(c)`

So, f is continuous at all points x ≠ 0

Case II:

` " If c = 0, then  " f(0)=0`

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^-} \left( x^2 \sin \frac{1}{x} \right) = \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right)\]
\[\text{It is known that} - 1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 . \]
\[ \Rightarrow - x^2 \leq x^2 \sin \frac{1}{x} \leq x^2 \]
\[ \Rightarrow \lim_{x \to 0} \left( - x^2 \right) \leq \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) \leq \lim_{x \to 0} x^2 \]
\[ \Rightarrow 0 \leq \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) \leq 0\]
\[ \Rightarrow \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) = 0\]
\[ \Rightarrow \lim_{x \to 0^-} f\left( x \right) = 0\]
\[\text{ Similarly } , \lim_{x \to 0^+} f\left( x \right) = \lim_{x \to 0^+} \left( x^2 \sin \frac{1}{x} \right) = \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) = 0\]

`∴ lim _(x->0^- ) f(x) = f(0) = lim_(x ->0^+ ) f (x)`

So, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.