Determine if f defined by: f(x)= x^ 2 1 x , if x 0 0, if x = 0

Question

Determine if f defined by:
$\text{f(x)}=\begin{cases}\text{x}^{2} \sin\frac{1}{\text{x}}, \text{if} \ \text{x}\neq0\\0, \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$

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Answer

It is given that $\text{f(x)}=\begin{cases}\text{x}^{2} \sin\frac{1}{\text{x}}, \text{if} \ \text{x}\neq0\\0, \ \ \ \ \ \ \ \ \ \ \ \text{if}\ \text{x} = 0\end{cases}$
We know that f is defined at all point of the real line.
Let k be a real number.
Case I: $\text{k} \neq 0,$
Then $\text{f(k)} =\text{k}^{2} \sin\frac{1}{\text{k}}$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) = \text{k}^{2}\sin \frac{\text{1}}{\text{k}}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$
Thus, f is continuous at all points x that is $\text{x}\neq0.$
Case II: k = 0
Then f(k) = f(0) = 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\Big( \text{x}^{2}\sin \frac{\text{1}}{\text{x}}\Big)$
We know that $-1 \leq\sin\frac{1}{\text {x}}\leq1, \text{x}\neq0$
$\rightarrow \text{x}^{2}\leq \text{x}^{2}\sin\frac{1}{\text {x}}\leq0$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) = 0$
$\Rightarrow^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = 0$
similarly,
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\Big(\text{x}^{2}\sin\frac{1}{\text{x}}\Big) =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}}\Big( \text{x}^{2}\sin \frac{\text{1}}{\text{x}}\Big) = 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = \text{f(0)}= ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)}$
Therefore , f is continuous at x = 0.
Therefore, f is has no point of discontinuity.

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