Show that the following functions have a continuous extension to …

Question

Show that the following functions have a continuous extension to the point where $f(x)$ is not defined. Also, find the extension:$f(x)=\frac{3 \sin ^2 x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^2 x\right)}$, for $x \neq 0$.

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Answer

$f(x)=\frac{3 \sin ^2 x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^2 x\right)}, \text { for } x \neq 0$
Here, $f(0)$ is not defined.
Consider,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{3 \sin ^2 x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^2 x\right)}$
$=\lim _{x \rightarrow 0} \frac{3 \sin ^2 x+2 \cos x \cdot\left(2 \sin ^2 x\right)}{2 \sin ^2 x}$
$=\lim _{x \rightarrow 0} \frac{\sin ^2 x(3+4 \cos x)}{2 \sin ^2 x}$
$=\lim _{x \rightarrow 0} \frac{3+4 \cos x}{2}$
$\cdots\left[\begin{array}{l}
\because x \rightarrow 0, x \neq 0, \sin x \neq 0 \\
\therefore \sin ^2 x \neq 0
\end{array}\right] \\
=\frac{1}{2} \lim _{x \rightarrow 0}(3+4 \cos x)=\frac{1}{3}(3+4 \cos 0)$
$=\frac{1}{2}(3+4)=\frac{7}{2}$
$\lim _{x \rightarrow 0} f(x) \text { exists. }$
But $f (0)$ is not defined.
$\therefore f ( x )$ has a removable discontinuity at $x =0$.
$\therefore$ The extension of the original function is
$f(x)=\frac{3 \sin ^2 x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^2 x\right)}, x \neq 0$
$=\frac{7}{2}, x=0$
$\therefore f(x)$ is continuous at $x=0$
$\therefore f ( x )$ is continuous at $x =0$.

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