Question

limx=0

$\sqrt{1 + x} - \sqrt{1 - x} \: divide \: by \:2 x$
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Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{2x}$

Let,

$\longrightarrow x=\cos(2\theta)$

As $x\to0,$

$\longrightarrow \cos(2\theta)\to0$

$\longrightarrow 2\theta\to\cos^{-1}(0)$

$\longrightarrow 2\theta\to\dfrac{\pi}{2}$

$\longrightarrow \theta\to\dfrac{\pi}{4}$

Then the limit becomes,

$\displaystyle\longrightarrow L=\dfrac{1}{2}\cdot\lim_{\theta\to\frac{\pi}{4}}\dfrac{\sqrt{1+\cos(2\theta)}-\sqrt{1-\cos(2\theta)}}{\cos(2\theta)}$

We know that,

• $1+\cos(2\theta)=2\cos^2\theta$

• $1-\cos(2\theta)=2\sin^2\theta$

• $\cos(2\theta)=\cos^2\theta-\sin^2\theta$

Then,

$\displaystyle\longrightarrow L=\dfrac{1}{2}\cdot\lim_{\theta\to\frac{\pi}{4}}\dfrac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\cos^2\theta-\sin^2\theta}$

$\displaystyle\longrightarrow L=\dfrac{1}{2}\cdot\lim_{\theta\to\frac{\pi}{4}}\dfrac{\sqrt2\,\cos\theta-\sqrt2\,\sin\theta}{\cos^2\theta-\sin^2\theta}$

$\displaystyle\longrightarrow L=\dfrac{1}{2}\cdot\lim_{\theta\to\frac{\pi}{4}}\dfrac{\sqrt2\,(\cos\theta-\sin\theta)}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}$

$\displaystyle\longrightarrow L=\dfrac{\sqrt2}{2}\cdot\lim_{\theta\to\frac{\pi}{4}}\dfrac{\cos\theta-\sin\theta}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}$

$\displaystyle\longrightarrow L=\dfrac{1}{\sqrt2}\cdot\lim_{\theta\to\frac{\pi}{4}}\dfrac{1}{\cos\theta+\sin\theta}$

Taking $\theta=\dfrac{\pi}{4},$

$\displaystyle\longrightarrow L=\dfrac{1}{\sqrt2}\cdot\dfrac{1}{\cos\left(\dfrac{\pi}{4}\right)+\sin\left(\dfrac{\pi}{4}\right)}$

$\displaystyle\longrightarrow L=\dfrac{1}{\sqrt2}\cdot\dfrac{1}{\left(\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}\right)}$

$\displaystyle\longrightarrow L=\dfrac{1}{\sqrt2}\cdot\dfrac{1}{\left(\dfrac{2}{\sqrt2}\right)}$

$\displaystyle\longrightarrow L=\dfrac{1}{\sqrt2}\cdot\dfrac{1}{\sqrt2}$

$\displaystyle\longrightarrow\underline{\underline{L=\dfrac{1}{2}}}}$

Hence 1/2 is the answer.