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Write the Value of Lim X → 0 √ 1 − Cos 2 X X .

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प्रश्न

Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]

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उत्तर

 
\[\lim_{x \to 0} \left( \frac{\sqrt{1 - \cos 2x}}{x} \right)\]
\[ = \lim_{x \to 0} \left( \frac{\sqrt{2 \sin^2 x}}{x} \right)\]
\[ = \sqrt{2} \lim_{x \to 0} \left( \frac{\sqrt{\sin^2 x}}{x} \right)\]
\[ = \sqrt{2} \lim_{x \to 0} \left( \frac{\left| \sin x \right|}{x} \right)\]
\[\text{ LHL }: \]
\[ = \sqrt{2} \lim_{x \to 0^-} \left( \frac{\left| \sin x \right|}{x} \right)\]
\[\text{ Let } x = 0 - h, \text{ where } h \to 0 . \]
\[ = \sqrt{2} \lim_{h \to 0} \left( \frac{\left| \sin \left( - h \right) \right|}{- h} \right)\]
\[ = \sqrt{2} \lim_{h \to 0} \left( \frac{\sin h}{- h} \right)\]
\[ = - \sqrt{2}\]
\[\text{ RHL }: \]
\[ = \sqrt{2} \lim_{x \to 0^+} \left( \frac{\left| \sin x \right|}{x} \right)\]
\[\text{ Let } x = 0 + h, \text{ where } h \to 0 . \]
\[ = \sqrt{2} \lim_{h \to 0} \frac{\left| \sin h \right|}{h} = \sqrt{2}\]
\[\text{ LHL } \neq RHL\]
\[\text{ Thus }, \lim_{x \to 0} \left( \frac{\sqrt{1 - \cos 2x}}{x} \right) \text{ does not exist } .\]

 

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अध्याय 29: Limits - Exercise 29.12 [पृष्ठ ७७]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.12 | Q 1 | पृष्ठ ७७

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