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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Determine if $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$ is a continuous function?
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Answer

The given function f is $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$
It is evident that f is defind at all points of the real line.
Let c be a real number.
Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\text{c}^2\sin\frac{1}{\text{c}}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\Big(\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\frac{1}{\text{x}}\Big)=\text{c}^2\sin\frac{1}{\text{c}}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$
So, f is continuouse at all points $\text{x}\neq0$
Case II:
If c = 0, then f(0) = 0
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow0^-}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)$
It is know that $-1\leq\sin\frac{1}{\text{x}}\leq1,\text{x}\neq0$
$\Rightarrow-\text{x}^2\leq\text{x}^2\sin\frac{1}{\text{x}}\leq\text{x}^2$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}(-\text{x}^2)\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq\lim\limits_{{\text{x}}\rightarrow0}\text{x}^2$
$\Rightarrow0\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=0$
Similarly, $\lim\limits_{{\text{x}}\rightarrow0^+}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}$
So, f is continuous at x = 0
From the above observations, it can be continuouse that f is continuous at every point of the real line.
Thus, f is a continuous function.

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