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COMEDK 2025 Afternoon Solved Paper

Question : 155 of 180

Marks: +1, -0

\lim_{\theta \to \frac{\pi}{2}} \frac{1-\sin \theta}{\left(\frac{\pi}{2}-\theta\right) \cos \theta}

is equal to :

Solution:

Concept:

Evaluate the limit using substitution and series expansion or standard trigonometric limits.

Explanation:

Let

h = \frac{\pi}{2} - \theta

. As

\theta \to \frac{\pi}{2}

h \to 0

Then

\sin \theta = \sin\left(\frac{\pi}{2} - h\right) = \cos h

and

\cos \theta = \cos\left(\frac{\pi}{2} - h\right) = \sin h

The limit becomes

\lim_{h\to 0} \frac{1 - \cos h}{h \cdot \sin h}

Using the series expansions:

1 - \cos h = \frac{h^2}{2} + O(h^4)

and

\sin h = h + O(h^3)

, the denominator is

h \cdot (h + O(h^3)) = h^2 + O(h^4)

Thus the limit is

\lim_{h\to 0} \frac{\frac{h^2}{2} + O(h^4)}{h^2 + O(h^4)} = \frac{1}{2}

Alternatively, use the standard limit

\lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}

and

\lim_{x\to 0}\frac{\sin x}{x}=1

, giving the same result.

Answer:

Option D,

\displaystyle\frac{1}{2}

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