\documentclass[12pt]{exam}
\noprintanswers
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\usepackage{amsmath}
\usepackage{graphicx,float}
\usepackage{wrapfig}
\usepackage{enumerate}

\usepackage{tikz}
\usepackage{pgfplots}

\begin{document}

\begin{questions}

\question % workshop bank Ch. 2 -> esergel25
Suppose $f(x) = \frac{ \sqrt{2 - \sqrt{4-x^2}} }{x}$.

\begin{parts}

\part Sketch the graph of $y=f(x)$ in the viewing window $[-2,2] \times [-1,1]$.

\part Based on the graph, what do you think the values of $\displaystyle \lim_{x \to 0^{-}} f(x)$, $\displaystyle \lim_{x \to 0^{+}} f(x)$, and $\displaystyle \lim_{x \to 0} f(x)$ should be?


\part Consider the following argument: We see algebraically that
$$ f(x) = \frac{ \sqrt{2 - \sqrt{4-x^2}} }{x} \cdot \frac{ \sqrt{2 + \sqrt{4-x^2}} }{ \sqrt{2 + \sqrt{4-x^2}} } = \frac{1}{ \sqrt{2 + \sqrt{4-x^2}} }. $$
Therefore
$$
\displaystyle \lim_{x \to 0} f(x) = f(0) = \frac{1}{2}.
$$

Does this agree with your prediction? Which do you think is wrong, the graph or the argument? Either way, find the reason for the discrepancy.

\end{parts}


\end{questions}

\end{document}