# Thread: limit Problem

1. ## limit Problem

$lim \frac{1+cos{\pi}x}{{tan^2}5{\pi}x} (x \to 1)$

2. See attachment

3. Thank's man , you think there is another solution without Lupital

4. I would be interested if there were

5. I post this if I find another solution.

6. Originally Posted by gilyos
Thank's man , you think there is another solution without Lupital
Sure. How about

$
\frac{1 + \cos \pi x}{\tan^2 5 \pi x}
$

$
=
(1 + \cos \pi x) \frac{ \cos^2 5 \pi x}{\sin^2 5 \pi x}
$

$
=
2(1 + \cos \pi x) \frac{ \cos^2 5 \pi x}{1 - \cos 10 \pi x}
$

$
=
2\cos^2 5 \pi x (1 + \cos \pi x) \cdot \frac{ 1 - \cos \pi x}{1 - \cos \pi x}\cdot \frac{ 1}{1 - \cos 10 \pi x} \cdot \frac{ 1 + \cos 10 \pi x}{1 + \cos 10 \pi x}
$

$
=
2\cos^2 5 \pi x \cdot \frac{ \sin^2 \pi x}{1 - \cos \pi x}\cdot \frac{ 1 + \cos 10 \pi x}{ \sin^2 10 \pi x}$

so

$
\lim_{x \to 1}
2\cos^2 5 \pi x \cdot \left(\frac{ \sin \pi x}{x}\right)^2 \cdot \left( \frac{x}{\sin^2 10 \pi x}\right)^2\cdot \frac{ 1 + \cos 10 \pi x}{1 - \cos \pi x} = 2(-1)^2 \frac{\pi^2}{100 \pi^2} \cdot \frac{2}{2} = \frac{1}{50}$

7. $\lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{x\to \pi}\frac{1+\cos x}{\tan^2 5x}=\lim_{t\to 0}\frac{1+\cos \left(\pi-t\right)}{\tan^2 5\left(\pi-t\right)}=\lim_{t\to 0}\frac{1-\cos t}{\tan^25t}$

so,

$\lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{t\to 0}\frac{1-\cos t}{t^2}\cdot\frac{25t^2}{\sin^2 5t}\cdot\frac{\cos^2 5t}{25}=\frac{1}{2}\cdot 1 \cdot\frac{1}{25}=\frac{1}{50}.$

8. Originally Posted by Abu-Khalil
$\lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{x\to \pi}\frac{1+\cos x}{\tan^2 5x}=\lim_{t\to 0}\frac{1+\cos \left(\pi-t\right)}{\tan^2 5\left(\pi-t\right)}=\lim_{t\to 0}\frac{1-\cos t}{\tan^25t}$

so,

$\lim_{x\to 1}\frac{1+\cos\pi x}{\tan^2 5\pi x}=\lim_{t\to 0}\frac{1-\cos t}{t^2}\cdot\frac{25t^2}{\sin^2 5t}\cdot\frac{\cos^2 5t}{25}=\frac{1}{2}\cdot 1 \cdot\frac{1}{25}=\frac{1}{50}.$
The scale and shift to the origin was a nice touch.

9. Good one !!!

10. Originally Posted by gilyos
$lim \frac{1+cos{\pi}x}{{tan^2}5{\pi}x} (x \to 1)$
I would have done the cool manipulation way, but it seems that's already been amply covered.

Let $L$ denote the above limit, then $L=\lim_{x\to0}\frac{1+\cos\left(\pi x-\pi\right)}{\tan^2\left(5\pi x-5\pi\right)}=\frac{1-\cos\left(\pi x\right)}{\tan^2\left(5\pi x\right)}$. Let $z=\pi x$ so our limit becomes $\lim_{z\to0}\frac{1-\cos(z)}{\tan^2(5z)}$ and it is well known that $\cos(z)\underset{z\to0}{\sim} 1-\frac{z^2}{2}$ and $\tan(z)\underset{z\to0}{\sim} z$ were it follows that $L=\lim_{z\to0}\frac{1-\left(1-\tfrac{z^2}{2}\right)}{25z^2}=\frac{1}{50}$