Question
Determine whether $\text{f(x)}=\begin{cases}\frac{\sin\text{x}^2}{\text{x}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$ is continuous at x = 0 or not.
✓
Answer
Given, $\text{f(x)}=\begin{cases}\frac{\sin\text{x}^2}{\text{x}},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
We have
$\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}}$
$=\lim\limits_{{\text{x}}\rightarrow0}\frac{\text{x}\sin\text{x}^2}{\text{x}^2}$
$=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}\lim\limits_{{\text{x}}\rightarrow0}\text{x}$
$=1\times0$
$=0$
$=\text{f}(0)$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0}\text{f(x)}=\text{f}(0)$
Hence, f(x) is continuous at x = 0.
We have
$\lim\limits_{{\text{x}}\rightarrow0}\text{f(x})=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}}$
$=\lim\limits_{{\text{x}}\rightarrow0}\frac{\text{x}\sin\text{x}^2}{\text{x}^2}$
$=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}\lim\limits_{{\text{x}}\rightarrow0}\text{x}$
$=1\times0$
$=0$
$=\text{f}(0)$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0}\text{f(x)}=\text{f}(0)$
Hence, f(x) is continuous at x = 0.
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