The question is given in the following picture:
My first answer to the question was A. My justification was as follows: $A(r) = \pi - \pi r^2$ and \begin{equation*} \lim_{ r \rightarrow 1^-} \frac{A(r)}{a(r)}= \lim_{ r \rightarrow 1^-}\frac{\pi - \pi r^2}{a(r)} \end{equation*}
And I substituted for $ r = 1$ in the numerator, so the numerator becomes 0, and this is for any value of the denominator then the answer is A as I choose.
But the right answer is E, so I remember that the area of the circle with center O contracts, but why we should know that when $r=1$ the area of the circle with center O should be 0, could anyone explain this for me?
When evaluating limits you need to be more careful. Consider:
\begin{equation*} \lim_{ x \rightarrow 0} \frac{x}{x}= 1 \end{equation*}
Although the numerator evaluates to 0 when $x = 0$ it should be clear that the limit isn't 0.
You should show that $a(r) = \pi(1-r)^2$. Then your limit becomes:
\begin{equation*} \lim_{ r \rightarrow 1^-} \frac{A(r)}{a(r)}= \lim_{ r \rightarrow 1^-}\frac{\pi - \pi r^2}{\pi(1-r)^2} = \lim_{ r \rightarrow 1^-}\frac{1 - r^2}{(1-r)^2} \end{equation*}
Can you do some algebra to simplify this?