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Mathematics Question on limits of trigonometric functions

The value of lim⁡x→01−cos x21−cos x\displaystyle \lim_{x \to 0} \frac{\sqrt{1-cos\,x^{2}}}{1-cos\,x} is

A

12\frac{1}{2}

B

22

C

2\sqrt{2}

D

None of these

Answer

2\sqrt{2}

Explanation

Solution

lim⁡x→01−cos x21−cos x=lim⁡x→02 sin2x222 sin2x2 \displaystyle\lim_{x \to 0} \frac{\sqrt{1-cos\,x^{2}}}{1-cos\,x} = \displaystyle\lim _{x \to 0} \frac{\sqrt{2\, sin^{2} \frac{x^{2}}{2}}}{2\, sin^{2} \frac{x}{2}} =lim⁡x→02∣sinx22∣2 sin2x2= \displaystyle\lim _{x \to 0} \frac{\sqrt{2} \left|sin \frac{x^{2}}{2}\right|}{2\, sin^{2} \frac{x}{2}} LHL=lim⁡x→0−2∣sinx22∣2 sin2x2=lim⁡h→02∣sin(−h)22∣2 sin2(−h)2LHL = \displaystyle\lim _{x \to 0^{-}} \frac{\sqrt{2} \left|sin \frac{x^{2}}{2}\right|}{2 \,sin^{2} \frac{x}{2}} = \displaystyle\lim _{h \to 0} \frac{\sqrt{2} \left|sin \frac{\left(-h\right)^{2}}{2}\right|}{2\, sin^{2} \frac{\left(-h\right)}{2}} lim⁡h→02 sinh222 sin2h2=22lim⁡h→0sin h22h22×(h2)2(sinh2)2×2\displaystyle\lim _{h \to 0} \frac{\sqrt{2}\, sin \frac{h^{2}}{2}}{2 \,sin^{2} \frac{h}{2}}= \frac{\sqrt{2}}{2} \displaystyle\lim _{h \to 0} \frac{ sin\, \frac{h^{2}}{2}}{ \frac{h^{2}}{2}}\times\frac{\left(\frac{h}{2}\right)^{2}}{\left(sin \frac{h}{2}\right)^{2}}\times2 [Note that the question contains mod sign, hence we checked for LHL and RHL]