# limit as x approaches pi/2 from the right side using l'Hospitals's Rule

• Oct 15th 2012, 02:26 PM
amthomasjr
limit as x approaches pi/2 from the right side using l'Hospitals's Rule
$\displaystyle \lim_{x \to(\pi/2)^+}\frac{cosx}{1-sinx}$

I found that:

$\displaystyle \lim_{x \to(\pi/2)^+}cosx=0$
$\displaystyle \lim_{x \to(\pi/2)^+}1-sinx=0$

so I found the derivatives of them and I got:
$\displaystyle \lim_{x \to(\pi/2)^+}\frac{sinx}{cosx}=\frac{sin(\pi/2)}{cos(\pi/2)}=\frac{1}{0}$

I'm not sure on what I am doing wrong.
• Oct 15th 2012, 02:50 PM
MaxJasper
Re: limit as x approaches pi/2 from the right side using l'Hospitals's Rule
lim from right -> $\displaystyle -\infty$
• Oct 15th 2012, 02:59 PM
HallsofIvy
Re: limit as x approaches pi/2 from the right side using l'Hospitals's Rule
Quote:

Originally Posted by amthomasjr
$\displaystyle \lim_{x \to(\pi/2)^+}\frac{cosx}{1-sinx}$

I found that:

$\displaystyle \lim_{x \to(\pi/2)^+}cosx=0$
$\displaystyle \lim_{x \to(\pi/2)^+}1-sinx=0$

so I found the derivatives of them and I got:
$\displaystyle \lim_{x \to(\pi/2)^+}\frac{sinx}{cosx}=\frac{sin(\pi/2)}{cos(\pi/2)}=\frac{1}{0}$

I'm not sure on what I am doing wrong.

You haven't, technically, done anything wrong, except that, of course, because you cannot divide by 0, you cannot evaluate that final limit by simply setting $\displaystyle x= \pi/2$. What you can do now is argue that, for x close to $\displaystyle \pi/2$, but larger, sin(x) will be very close to 1 while cos(x) will be very close to 0 and negative. That gives the result that the limit is "$\displaystyle -\infty$" as MaxJasper said. I will add that "$\displaystyle -\infty$" is not a real number so that is just saying "the limit does not exist", in a particular way.
• Oct 15th 2012, 04:08 PM
amthomasjr
Re: limit as x approaches pi/2 from the right side using l'Hospitals's Rule
Thanks