Question

${lim}_{x\to o}cos\left(\sqrt{1+x}\right)-cos\sqrt{x}$.

Answer 1

We have,

${lim}_{x\to 0}\cos \left(\sqrt{1+x}\right)-\cos \sqrt{x}$

$\Rightarrow {lim}_{x\to 0}2sin\frac{(\sqrt{1+x}+\sqrt{x})}{2}sin\frac{(\sqrt{x}-\sqrt{1+x})}{2}\therefore \cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}$

$\Rightarrow {lim}_{x\to 0}2sin\frac{(\sqrt{1+x}+\sqrt{x})}{2}sin\frac{(\sqrt{x}-\sqrt{1+x})}{2}\times \frac{(\sqrt{x}+\sqrt{1+x})}{(\sqrt{x}+\sqrt{1+x})}\text{on rationlize}$

$\Rightarrow {lim}_{x\to 0}2sin\frac{(\sqrt{1+x}+\sqrt{x})}{2}sin\frac{(x-1-x)}{2(\sqrt{x}+\sqrt{1+x})}$

Taking limit and we get,

$\Rightarrow 2\sin \frac{(\sqrt{1+0}+\sqrt{0})}{2}\sin \frac{-1}{2(\sqrt{0}+\sqrt{1+0})}$

$\Rightarrow 2\sin \frac{1}{2}\sin \frac{-1}{2}$

$\Rightarrow -2{\sin }^2\frac{1}{2}$

Hence, this is the answer