Solving the Limit Problem of Death: A Tricky Mathematical Challenge

[danni7070]

[Solved] Limit problem of death

$$\lim_{x\rightarrow{\pi/2}} (x*tan(x) - \frac{\pi}{2*cos(x)})$$

OK. I'm not sure how to begin this problem because:

1) tan(pi/2) is undefined and

2) pi/2cos(pi/2) is undefined also!

Any hint would be great!

Thanks

 

[Dick]

Combine them into a single fraction and apply l'Hopital.

 

[danni7070]

like this?

$$\frac{2cos(x)*x*tan(x)-\pi}{2cos(x)}$$

This looks incorrect...

 

[cristo]

You can write tanx=sinx/cosx, so try to do that with your expression in the first post.

 

[Dick]

You can write tanx=sinx/cosx, so try to do that with your expression in the first post.

Or just put it into your current version. cos(x)*tan(x)=sin(x).

 

[cristo]

Or just put it into your current version. cos(x)*tan(x)=sin(x).

yup, that works as well, and is a lot simpler to do!

 

[danni7070]

Ok, I'm still confused. I must be overseeing something obvious.

I get

$$\frac{xsin(x)- \pi}{cos{x}}$$

That doesn't work for l'Hôpital so I must be doing something wrong.

 

[Dick]

What happened to the '2's? It should be a 0/0 limit.

 

[danni7070]

This is what I did

$$\frac{2cos(x)*x*\frac{sinx}{cosx}-\pi}{2cos(x)}$$

$$\frac{2cos(x)*x*sin(x)-\pi*cos(x)}{cos(x)} * \frac{1}{2cos(x)}$$

$$\frac{2cos(x)*x*sin(x)-\pi*cos(x)}{2cos^2(x)}$$

Another result and they are both 0/0 but is this correct ?

 

[danni7070]

Ok I think I got it! The above equals to

$$\frac{x*sin(x)-\pi/2}{cos(x)} : [\frac{0}{0}]$$

 

[danni7070]

Ok thanks a bunch guys. The rest is simple

$$\lim_{x\rightarrow\pi/2} \frac{sin(x)+x*cos(x)}{-sin(x)} = \frac{\pi/2}{-\pi/2} = -1$$

Edit: It is not pi/2 there it is supposed to be 1/-1