Question

How is lim x-> pi/2 from the right = -infinity?

Answer 1

\[\lim_{x \rightarrow \frac{ \pi }{ 2 }^+}\frac{ 1 }{ 0 }=-\infty \]

Answer 2

The original problem is this:\[\lim_{x \rightarrow \frac{ \pi }{ 2 }^+}\frac{ cosx }{ 1-sinx }\]