Answer
(a) $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$
(b) As $x\to0$, all of the three graphs approach $y=\frac{1}{2}$, squeezing together as they go.
Work Step by Step
(a) - Calculate $\lim_{x\to0}(\frac{1}{2}-\frac{x^2}{24})$ and $\lim_{x\to0}(\frac{1}{2})$
$\lim_{x\to0}\Big(\frac{1}{2}-\frac{x^2}{24}\Big)=\frac{1}{2}-\frac{0^2}{24}=\frac{1}{2}-0=\frac{1}{2}$
$\lim_{x\to0}\frac{1}{2}=\frac{1}{2}$
So, $\lim_{x\to0}(\frac{1}{2}-\frac{x^2}{24})=\lim_{x\to0}\frac{1}{2}=\frac{1}{2}$
- Yet $\frac{1}{2}-\frac{x^2}{24}\le \frac{1-\cos x}{x^2}\le \frac{1}{2}$ for all $x$ close to $0$
Therefore, applying the Sandwich Theorem, we can conclude that $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$
A note here is that the inequality does not have to hold for $x=0$ though, if it already holds for all values of $x$ close to $0$ for the simple reason that $\lim_{x\to0}$ accounts for only the values of $x$ close to $0$ and does not account for $0$.
(b) The graph is shown below.
As $x\to0$, all of the three graphs approach $y=\frac{1}{2}$, squeezing together as they go.
