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NCERT Exemplar · Q25

Q.Evaluate lim⁡x→π6cot⁡2x−3cosec⁡x−2\lim_{x \to \frac{\pi}{6}} \dfrac{\cot^2 x - 3}{\operatorname{cosec} x - 2}.

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Substitution gives 00\frac{0}{0}. Use cot⁡2x=cosec⁡2x−1\cot^2 x=\operatorname{cosec}^2 x-1 so the numerator becomes a difference of squares that cancels the denominator, leaving cosec⁡x+2→4\operatorname{cosec} x+2\to4.

Step 1 — Check the form. At x=π6x=\frac{\pi}{6}, cot⁡π6=3\cot\frac{\pi}{6}=\sqrt{3} and cosec⁡π6=2\operatorname{cosec}\frac{\pi}{6}=2, so the numerator is (3)2−3=0(\sqrt{3})^2-3=0 and the denominator is 2−2=02-2=0: the form 00\frac{0}{0}.

Step 2 — Rewrite the numerator. Using the identity cot⁡2x=cosec⁡2x−1\cot^2 x=\operatorname{cosec}^2 x-1,

cot⁡2x−3=cosec⁡2x−1−3=cosec⁡2x−4=(cosec⁡x−2)(cosec⁡x+2).\cot^2 x-3=\operatorname{cosec}^2 x-1-3=\operatorname{cosec}^2 x-4=(\operatorname{cosec} x-2)(\operatorname{cosec} x+2).

Step 3 — Cancel the common factor. …

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