Limit to 0

Let $f(x) = \displaystyle \int_0^x \sqrt{1+\cos(t)}\,dt$ . Then we have that the limit is in the form $\lim_{x\to 0} \frac{f(x)-f(0)}{x-0} \Rightarrow \lim_{x\to a}\frac{f(x)-f(a)}{x-a} = f'(a).$ So we can easily evaluate the limit $\begin{aligned} f'(0) = & \lim_{x\to0}\frac{f(x)-f(0)}{x-0}\\ =& \frac{d}{dx}\Big|_{x=0} \int_0^x \sqrt{1+\cos(t)}\,dt \\ = & \sqrt{1+\cos(x)}\Big|_{x=0} \\ = & \boxed{\sqrt{2}} \end{aligned}$

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 0} \frac {\displaystyle \int_0^x \sqrt{1+\cos t}\ dt}x & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\sqrt{1+\cos x}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x. \\ & = \boxed{\sqrt 2} \end{aligned}$

$\lim_{x\to0}\frac{\int_0^x \sqrt{1+\cos t}dt}{x}$ = $\lim_{x\to0}\frac{\int_0^x \sqrt{\cos^2 (\frac{1}{2} t) + \sin^2 (\frac{1}{2} t)+\cos t}dt}{x}$ = $\lim_{x\to0}\frac{\int_0^x \sqrt{\cos^2 (\frac{1}{2} t) + \sin^2 (\frac{1}{2} t)+\cos^2 (\frac{1}{2} t) - \sin^2 (\frac{1}{2} t)}dt}{x}$ = $\lim_{x\to0}\frac{\int_0^x \sqrt{2\cos^2(\frac{1}{2}t)}dt}{x}$ = $\lim_{x\to0}\frac{\int_0^x \sqrt{2}\cos(\frac{1}{2}t)dt}{x}$ = $\lim_{x\to0}\frac{2\sqrt{2}\sin(\frac{1}{2}t)|_0^x}{x}$ = $\lim_{x\to0}\frac{2\sqrt{2}\sin(\frac{1}{2}x)}{x}$ = $2\sqrt{2}\times\frac{1}{2}$ = $\sqrt{2}$