Limits, Continuity & Differentiability
Limits of Trigonometric Expressions
nta_pyq_2025_apr
Grade 12

Question:

$\lim_{x \to 0} \csc x\!\left(\sqrt{2\cos^2 x + 3\cos x} - \sqrt{\cos^2 x + \sin x + 4}\right)$ is:
0
$\frac{1}{\sqrt{15}}$
$\frac{1}{2\sqrt{5}}$
$-\frac{1}{2\sqrt{5}}$

Step-by-Step Solution

Key Concept: Multiply numerator and denominator by the conjugate of the square root difference to rationalize, then evaluate the limit using $\lim_{x\to0}\sin x/x = 1$.
Rationalize: numerator $\to(\cos^2x+3\cos x-\sin x-4)$. Using half-angle: numerator $\to -2\sin^2(x/2)(\cos x+4)-2\sin(x/2)\cos(x/2)$. After dividing by $\sin x=2\sin(x/2)\cos(x/2)$, limit $\to -\frac{1}{2\sqrt{5}}$.
Correct Answer: $-\frac{1}{2\sqrt{5}}$

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