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The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$
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Verified Answer
The correct answer is:
does not exist
$$
\text { Let } \begin{array}{rlr}
L & =\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} & \\
& =\lim _{x \rightarrow 1} \frac{\sqrt{2 \sin ^2(x-1)}}{x-1} \quad\left[\because 1-\cos 2 \theta=2 \sin ^2 \theta\right] \\
& =\sqrt{2} \lim _{x \rightarrow 1} \frac{|\sin (x-1)|}{x-1} & {\left[\frac{0}{0}\right. \text { form ] }} \\
& =\sqrt{2} \lim _{(x-1) \rightarrow 0} \frac{|\sin (x-1)|}{x-1} & \\
& =\sqrt{2} \lim _{z \rightarrow 0} \frac{|\sin z|}{z} & \text { [Let } x-1=z]
\end{array}
$$
[Let $x-1=z$ ]
$$
\begin{aligned}
& \text { NHL }=\sqrt{2} \lim _{z \rightarrow 0^{+}} \frac{\sin z}{z}=\sqrt{2} \\
& \text { LHL }=\sqrt{2} \lim _{z \rightarrow 0^{-}}-\frac{\sin z}{z}=-\sqrt{2}
\end{aligned}
$$
$\because$ RHL $\neq$ LH
$\therefore$ Limit does not exist.
\text { Let } \begin{array}{rlr}
L & =\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} & \\
& =\lim _{x \rightarrow 1} \frac{\sqrt{2 \sin ^2(x-1)}}{x-1} \quad\left[\because 1-\cos 2 \theta=2 \sin ^2 \theta\right] \\
& =\sqrt{2} \lim _{x \rightarrow 1} \frac{|\sin (x-1)|}{x-1} & {\left[\frac{0}{0}\right. \text { form ] }} \\
& =\sqrt{2} \lim _{(x-1) \rightarrow 0} \frac{|\sin (x-1)|}{x-1} & \\
& =\sqrt{2} \lim _{z \rightarrow 0} \frac{|\sin z|}{z} & \text { [Let } x-1=z]
\end{array}
$$
[Let $x-1=z$ ]
$$
\begin{aligned}
& \text { NHL }=\sqrt{2} \lim _{z \rightarrow 0^{+}} \frac{\sin z}{z}=\sqrt{2} \\
& \text { LHL }=\sqrt{2} \lim _{z \rightarrow 0^{-}}-\frac{\sin z}{z}=-\sqrt{2}
\end{aligned}
$$
$\because$ RHL $\neq$ LH
$\therefore$ Limit does not exist.
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