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

**amthomasjr** · October 15th 2012, 02:26 PM

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

I found that:

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

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

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

**MaxJasper** · October 15th 2012, 02:50 PM

lim from right -> $-\infty$

**HallsofIvy** · October 15th 2012, 02:59 PM

Originally Posted by amthomasjr

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

I found that:

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

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

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 $x= \pi/2$ . What you can do now is argue that, for x close to $\pi/2$ , 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 " $-\infty$ " as MaxJasper said. I will add that " $-\infty$ " is not a real number so that is just saying "the limit does not exist", in a particular way.

**amthomasjr** · October 15th 2012, 04:08 PM

Thanks

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