Q:

What is the limit of $\frac{1}{1-p}$ as $p \rightarrow 1$

What is the limit of $\frac{1}{1-p}$ as $p \rightarrow 1$. I know this is very simple limit question, but It has been a long while since I have done it. If you could include the steps then that would be great. 

A:

The limit does not exist,

in the strict sense: the function does not tend to any finite value;
even if you allow infinite limits: the left- and right-handed limits are different.

The numerator is constant, non-zero and clearly positive. The denominator tends to $0$ as $p \to 1$. The absolute value of a fraction with non-zero numerator and denominator tending to $0$, tends to infinity. You need to be careful with the sign though.
For values of $p$ smaller than 1, the denominator is positive, and so is the fraction; while for values of $p$ larger than 1, the denominator is negative, and so is the fraction. This behavior is captured by the one-sided limits:
$$\lim_{x \to 1^-} \frac{1}{1-p} =\lim_{x \underset{<}{\to} 1} \frac{1}{1-p}= \color{green}{+\infty} \quad \mbox{and} \quad \lim_{x \to 1^+} \frac{1}{1-p} =\lim_{x \underset{>}{\to} 1} \frac{1}{1-p}= \color{red}{-\infty}$$

Remark
If the signs would be the same on both sides, the function still wouldn't have a finite limit but you could say the limit is either $+\infty$ or $-\infty$; for example:
$$\lim_{x \to 1} \frac{1}{\left(1-p\right)^2} = +\infty$$

