Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
Which of the following limits vanish?
$\lim_{x \to 0^+} \sin \frac{x}{\sqrt{x}}$
$\lim_{x \to \pi/2} (1 - \sin x) \cdot \tan x$
$\lim_{x \to \infty} \frac{2x^2 + 3}{x^2 + x - 5} \cdot \sgn(x)$
$\lim_{x \to 3} \frac{[x]^2 - 9}{x^2 - 9}$
Step-by-Step Solution
Key Concept: Algebraic manipulation and trigonometric identities are essential to resolve indeterminate forms before evaluating limits.
The limit $\lim_{x \to 0} \frac{(1-\sin x)\sin x}{\cos x} \cdot \frac{1+\sin x}{1+\sin x}$ simplifies using the numerator $(1-\sin x)\sin x(1+\sin x) = \sin x(1-\sin^2 x) = \sin x \cos^2 x$. Thus the limit becomes $\lim_{x \to 0} \frac{\sin x \cos^2 x}{\cos x(1+\sin x)} = \lim_{x \to 0} \sin x \cos x = 0$. The second limit: $\lim_{x \to \pi/2} \frac{\sin x \cos x}{1+\sin x} = \frac{1 \cdot 0}{1+1} = 0$.
Correct Answer: 1,2,3