# Question #29a12

Feb 5, 2017

$f \left(x\right)$ is continuous at $x = 0$.

#### Explanation:

We say that a function $f \left(x\right)$ is continuous at ${x}_{0}$ if

• $f \left({x}_{0}\right)$ exists
• ${\lim}_{x \to {x}_{0}^{+}} f \left(x\right) = f \left({x}_{0}\right)$
• ${\lim}_{x \to {x}_{0}^{-}} f \left(x\right) = f \left({x}_{0}\right)$.

For the given $f \left(x\right)$ and ${x}_{0} = 0$, the first condition is satisfied as $f \left(0\right) = \frac{1}{2}$. To evaluate the limits, observe that $f \left(x\right)$ is identical to the function $\frac{1 - \cos \left(x\right)}{x} ^ 2$ at all points other than $0$, and thus we may work on that function when calculating the limit as $f \left(x\right)$ approaches $0$.

To evaluate ${\lim}_{x \to {0}^{+}} \frac{1 - \cos \left(x\right)}{x} ^ 2$, we note that $1 - \cos \left(x\right)$ and ${x}^{2}$ are differentiable on an open interval with the endpoint $0$, for example, $\left(0 , 1\right)$. Additionally, $\frac{d}{\mathrm{dx}} {x}^{2} \ne 0$ and direct substitution of $x = 0$ leads to an indeterminate form $\frac{0}{0}$. Thus we may apply L'Hopital's rule.

${\lim}_{x \to {0}^{+}} \frac{1 - \cos \left(x\right)}{x} ^ 2 = \frac{\frac{d}{\mathrm{dx}} \left(1 - \cos \left(x\right)\right)}{\frac{d}{\mathrm{dx}} {x}^{2}}$

$= {\lim}_{x \to {0}^{+}} \sin \frac{x}{2 x}$

$= \frac{1}{2} {\lim}_{x \to {0}^{+}} \sin \frac{x}{x}$

$= \frac{1}{2} \left(1\right)$

The above follows from the well known limit ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$. It may also be verified via a second application of L'Hopital's rule.

$= \frac{1}{2}$

$= f \left(0\right)$.

Verification of the limit ${\lim}_{x \to {0}^{-}} f \left(x\right) = \frac{1}{2}$ follows in much the same manner.

As all three conditions are satisfied, we can say that $f \left(x\right)$ is continuous at $x = 0$.