Question

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

Answer 1

Let we do this problem by mean value theorem

Let $g\left(x\right)=\cos \left(\sqrt{x}\right)$

By mean value theorem

$\Rightarrow g^{{}^{\prime }}\left(ϵ\right)=\frac{g(x+1)-g\left(x\right)}{x+1-x}$ where $x<ϵ<x+1$

$\Rightarrow \frac{\sin \left(\sqrt{x}\right)}{2\sqrt{x}}=\cos \left(\sqrt{x+1}\right)-\cos \left(\sqrt{x}\right)$

$\Rightarrow li{m}_{x\to \infty }\cos \left(\sqrt{x+1}\right)-\cos \left(\sqrt{x}\right)={lim}_{x\to \infty }\frac{\sin \left(\sqrt{x}\right)}{2\sqrt{x}}<li{m}_{x\to \infty }\frac{1}{2\sqrt{x}}=0$

Hence the answers to given limit is 0