lim (x->1) of (x-1-ln x)/(1+cos (pi*x) )

[csmajor86]

I need to evaluate the limit:

lim[sub:3i8t85qb]x->1[/sub:3i8t85qb] x -1 - ln x / 1 + cos(pi*x)

This is 0/0 form.

I used l'Hopital's rule once to get:

lim[sub:3i8t85qb]x->1[/sub:3i8t85qb] (1 - 1/x) / -sin(pi*x)*pi

which is still in 0/0 form

I then used l'Hopital's rule again to get:

lim[sub:3i8t85qb]x->1[/sub:3i8t85qb] (-1/x[sup:3i8t85qb]2[/sup:3i8t85qb])/ - cos(pi*x) pi[sup:3i8t85qb]2[/sup:3i8t85qb] = 1/pi[sup:3i8t85qb]2[/sup:3i8t85qb]

Is that right or did I miss something? For some reason feel like I'm missing something.

Thanks for your help.

 

[o_O]

When applying l'hopital the second time, you kept the negative sign in the numerator by accident:

\(\displaystyle \lim_{x \to 1} \left(\frac{1 - \frac{1}{x}}{-\pi sin \pi x}\right) \quad \stackrel{\mbox{l'Hopital}}{\longrightarrow} \quad \lim_{x \to 1} \left(\frac{\frac{1}{x^{2}}}{- \pi^{2} cos \pi x}\right)\)

You arrived at the right answe though. So other than that, it looks pretty good to me.