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Question: \(\lim_{x \rightarrow 0}\frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\) is equal to...

lim⁡x→0x1+x−1−x\lim_{x \rightarrow 0}\frac{x}{\sqrt{1 + x} - \sqrt{1 - x}} is equal to

A

12\frac{1}{2}

B

2

C

1

D

0

Answer

1

Explanation

Solution

lim⁡x→0(x1+x−1−x)=lim⁡x→0(x1+x−1−x×1+x+1−x1+x+1−x)\lim_{x \rightarrow 0}\left( \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}} \right) = \lim_{x \rightarrow 0}\left( \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}} \times \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right)

=lim⁡x→0(x(1+x+1−x)1+x−1+x)=lim⁡x→0((1+x+1−x)2)=22=1= \lim_{x \rightarrow 0}\left( \frac{x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)}{1 + x - 1 + x} \right) = \lim_{x \rightarrow 0}\left( \frac{\left( \sqrt{1 + x} + \sqrt{1 - x} \right)}{2} \right) = \frac{2}{2} = 1