Looks like the limit is
[tex]\displaystyle\lim_{x\to\pi/2^+}\frac{\cos x}{1-\sin x}[/tex]
which yields an indeterminate form [tex]\dfrac00[/tex]. Rewriting as
[tex]\dfrac{\cos x(1+\sinx)}{(1-\sin x)(1+\sin x)}=\dfrac{\cos x(1+\sin x)}{1-\sin^2x}=\dfrac{1+\sin x}{\cos x}[/tex]
we see the numerator approaches 1 + 1 = 2, while the denominator approaches 0. Since [tex]\cos x<0[/tex] for [tex]x[/tex] near [tex]\dfrac\pi2[/tex] with [tex]x>\dfrac\pi2[/tex], the limit is [tex]-\infty[/tex].
