The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$
  1. exists and is equal to $\sqrt{2}$
  2. exists and is equal to $-\sqrt{2}$
  3. does not exist
  4. exists and is equal to $\left(\frac{1}{2}\right)$

Solution

$ \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

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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