What is {θ →0}{lim}{{1-cosθ }}{θ } equal to?

Correct Answer is : 12{1}{{2}}21

Correct Answer is : 12{1}{{2}}21

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UPSC NDA II 2018 Mathematics Solved Paper

Question : 111 of 120

Marks: +1, -0

What is

\underset{\theta →0}{lim}\frac{\sqrt{1-cos\theta }}{\theta }

equal to?

Solution:

Concept:

Use the identity

1-\cos\theta = 2\sin^2\frac{\theta}{2}

and the standard limit

\underset{x\to0}{\lim}\frac{\sin x}{x}=1

Explanation:

Rewrite the numerator using the identity:

\sqrt{1-\cos\theta} = \sqrt{2\sin^2\frac{\theta}{2}} = \sqrt{2}\left|\sin\frac{\theta}{2}\right|

For

\theta\to0

\sin\frac{\theta}{2}\ge0

, so the absolute value can be dropped:

\sqrt{2}\sin\frac{\theta}{2}

Thus the limit becomes

\underset{\theta\to0}{\lim}\frac{\sqrt{2}\sin\frac{\theta}{2}}{\theta}

Write

\theta = 2\cdot\frac{\theta}{2}

\frac{\sqrt{2}\sin\frac{\theta}{2}}{2\cdot\frac{\theta}{2}} = \frac{\sqrt{2}}{2}\cdot\frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}}

\theta\to0

\frac{\theta}{2}\to0

, so

\frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}}\to1

Therefore the limit equals

\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

Answer:

\frac{1}{\sqrt{2}}

(Option C).

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