How to calculate $\lim\limits_{x\to1^+}\frac{1}{(x-1)^2} \int\limits_{1}^{x} \sqrt{1+\cos(\pi t)}\,\mathrm dt$ Can anyone help me by calculating this limit?
I know that I need L'Hôpital but how?
$$
\lim_{x \to 1^+}\frac{1}{(x-1)^2} \int_{1}^{x} \sqrt{1+\cos(\pi t)} \,\mathrm dt
$$
Thank you very much!!
 A: Since my hints were not helpful, here a is my solution
$$
\begin{align}
\lim\limits_{x\to 1^+}\frac{1}{(x-1)^2}\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt
&=\lim\limits_{x\to 1^+}\frac{\left(\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt\right)'}{((x-1)^2)'}&\text{L'Hopitale rule}\\
&=\lim\limits_{x\to 1^+}\frac{\sqrt{1+\cos(\pi x)}}{2(x-1)}& y\to x-1\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{1+\cos(\pi (y+1))}}{2y}&\cos(\pi+\alpha)=-\cos\alpha\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{1-\cos(\pi y)}}{2y}&1-\cos\alpha=2\sin^2\frac{\alpha}{2}\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{2}|\sin\frac{\pi y}{2}|}{2y}&y>0\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\sin\frac{\pi y}{2}}{2y}&\sin\alpha\sim\alpha\\
&=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\frac{\pi y}{2}}{2y}\\
&=\frac{\pi\sqrt{2}}{4}
\end{align}
$$
A: Integrate the equivalent, valid when $t\gt1$, $t\to1$,
$$
\sqrt{1+\cos(\pi t)}=\sqrt2\,\sin\left(\frac\pi2(t-1)\right)\sim\frac\pi{\sqrt2}\,(t-1).
$$
This yields, for $x\to1$, $x\gt1$,
$$
\int_1^x\sqrt{1+\cos(\pi t)}\mathrm dt\sim\frac\pi{\sqrt2}\int_1^x(t-1)\mathrm dt=\frac\pi{2\sqrt2}(x-1)^2.
$$
