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Model Paper 6 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 61 Mark
MCQ
$\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$ is equal to
  • A
    1
  • B
    $0$
  • C
    2
  • D
    -1
✓

Answer

(a) 1
Explanation: Given, $\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$
$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{\sin x[\sqrt{x+1}+\sqrt{1-x}]}{(\sqrt{x+1}-\sqrt{1-x}) \cdot(\sqrt{x+1}+\sqrt{1-x})} \\ =\lim _{x \rightarrow 0} \frac{\sin x[\sqrt{x+1}+\sqrt{1-x}]}{x+1-1+x} \\ =\lim _{x \rightarrow 0} \frac{\sin x[\sqrt{x+1}+\sqrt{1-x}]}{2 x}=\frac{1}{2} \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x}[\sqrt{x+1}+\sqrt{1-x}]\end{array}$
Taking limits, we get
$=\frac{1}{2} \times 1 \times[\sqrt{0+1}+\sqrt{1-0}]=\frac{1}{2} \times 1 \times 2=1$

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