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NCERT Exemplar · Q59

Q.lim⁡x→0cosec⁡x−cot⁡xx\lim_{x \to 0} \dfrac{\operatorname{cosec} x - \cot x}{x} is
(A) −12\dfrac{-1}{2}
(B) 11
(C) 12\dfrac{1}{2}
(D) 11

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Writing cosec⁡x−cot⁡x=1−cos⁡xsin⁡x\operatorname{cosec} x - \cot x = \frac{1-\cos x}{\sin x} and using 1−cos⁡x=2sin⁡2x21-\cos x = 2\sin^2\frac{x}{2} with lim⁡θ→0sin⁡θθ=1\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1 gives 12\frac{1}{2} — option (C).

Step 1 — Rewrite in sine and cosine.

cosec⁡x=1sin⁡x,cot⁡x=cos⁡xsin⁡x,\operatorname{cosec} x = \frac{1}{\sin x}, \qquad \cot x = \frac{\cos x}{\sin x},

so

cosec⁡x−cot⁡x=1sin⁡x−cos⁡xsin⁡x=1−cos⁡xsin⁡x.\operatorname{cosec} x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x}.

The limit becomes

lim⁡x→01−cos⁡xx sin⁡x.\lim_{x\to0}\frac{1-\cos x}{x\,\sin x}.

Step 2 — Replace 1−cos⁡x1-\cos x with a half-angle form.

Using 1−cos⁡x=2sin⁡2x21-\cos x = 2\sin^2\frac{x}{2} and sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}:

1−cos⁡xx sin⁡x=2sin⁡2x2x⋅2sin⁡x2cos⁡x2=sin⁡x2x cos⁡x2.\frac{1-\cos x}{x\,\sin x} = \frac{2\sin^2\frac{x}{2}}{x\cdot 2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{x\,\cos\frac{x}{2}}.

Step 3 — Shape it into sin⁡θθ\frac{\sin\theta}{\theta}.

Write sin⁡x2=x2⋅sin⁡x2x2\sin\frac{x}{2} = \frac{x}{2}\cdot\dfrac{\sin\frac{x}{2}}{\frac{x}{2}}: …

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