2
$\begingroup$

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!!

$\endgroup$
  • $\begingroup$ Please use different symbols for the upper bound of the integral and for the integrand. $\endgroup$ – Did May 21 '13 at 15:31
5
$\begingroup$

Since my hints were not helpful, here a is my solution $$ \begin{align} \lim\limits_{x\to 1^+}\frac{1}{(x-1)^2}\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt &=\lim\limits_{x\to 1^+}\frac{\left(\int\limits_{1}^{x}\sqrt{1+\cos(\pi t)}dt\right)'}{((x-1)^2)'}&\text{L'Hopitale rule}\\ &=\lim\limits_{x\to 1^+}\frac{\sqrt{1+\cos(\pi x)}}{2(x-1)}& y\to x-1\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{1+\cos(\pi (y+1))}}{2y}&\cos(\pi+\alpha)=-\cos\alpha\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{1-\cos(\pi y)}}{2y}&1-\cos\alpha=2\sin^2\frac{\alpha}{2}\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{2}|\sin\frac{\pi y}{2}|}{2y}&y>0\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\sin\frac{\pi y}{2}}{2y}&\sin\alpha\sim\alpha\\ &=\lim\limits_{y\to 0^+}\frac{\sqrt{2}\frac{\pi y}{2}}{2y}\\ &=\frac{\pi\sqrt{2}}{4} \end{align} $$

$\endgroup$
  • $\begingroup$ $$ \lim_{x,1+}\frac{1}{(x-1)^2} \int_{1}^{x} \sqrt{1+cos(\pi t)} dt $$ i used L'Hospital so i got this: $$\frac{\sqrt{1+cos(\pi *x)}} {2*(x-1)}$$ so i used l'Hospital again: and again...and again so i got this: $$\frac{\pi * (2*cos(\pi*x))+(3*cos(2*\pi *x)+2))*\sqrt{1+cos(\pi*x)}} {2*sin(\pi*x)*(3*cos(\pi*x)+2)}$$ now i dont know how to continue. $\endgroup$ – Xorkox May 21 '13 at 15:55
  • $\begingroup$ You need to aplly that rule only once, then substitute $y=x-1$, then $y\to 0$ $\endgroup$ – Norbert May 21 '13 at 16:04
  • $\begingroup$ @Xorkox Is it clear now? $\endgroup$ – Norbert May 21 '13 at 20:07
  • $\begingroup$ no :/ substitude this doesnt help me :( i am confused $\endgroup$ – Xorkox May 22 '13 at 7:19
  • $\begingroup$ @Xorkox see my answer $\endgroup$ – Norbert May 22 '13 at 10:43
1
$\begingroup$

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 $x\to1$, $x\gt1$, $$ \int_1^x\sqrt{1+\cos(\pi t)}\mathrm dt\sim\frac\pi{\sqrt2}\int_1^x(t-1)\mathrm dt=\frac\pi{2\sqrt2}(x-1)^2. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.