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BITSAT 2021

Exam: BITSAT
Year: 2021
Subject: Mathematics
Chapter: Limits

BITSAT 2021 · Mathematics · Limits

_ x arrow 0 x- x x+ ^ 2 x is equal to

\lim _{x \rightarrow 0} \sqrt{\frac{x-\sin x}{x+\sin ^{2} x}}

is equal to

Options

A1
B0
C $\infty$
DNone of these

Correct answer

B. 0

Step-by-step solution

\lim _{x \rightarrow 0} \sqrt{\frac{x-\sin x}{x+\sin ^{2} x}=\lim _{x \rightarrow 0} \sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{\sin ^{2} x}{x}}}}=\lim _{x \rightarrow 0} \sqrt{\frac{1-\frac{\sin x}{x}}{1+\left(\frac{\sin x}{x}\right) \sin x}}=\sqrt{\frac{1-1}{1+1 \times 0}}=0

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