91Ó°ÊÓ

Consider the function given as:

$f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos x}{x},ifx≠0\\ 0,ifx=0\end{array}\right.$

Derivative test states that,
If $f$is a function defined from $(a,b)→R$and $k∈(a,b)$, then $f$is differentiable at $k$if the limit $\underset{h→∞}{\lim }\frac{f(k+h)-f\left(h\right)}{h}$ exists and equals $f^{\text{'}}\left(k\right)$.

The objective is to use the derivative test to prove that the function $f$ is differentiable at $k=0$.

Prove the function $f$is differentiable at $k=0$as follows:

Consider the following expression and solve as,
$\begin{array}{rcl}\underset{h→∞}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}& =& \underset{h→∞}{\lim }\left[\frac{\frac{1-\cos h}{h}-0}{h-0}\right]\\ & =& \underset{h→∞}{\lim }\left[\frac{{\displaystyle 1-\cos h}}{{\displaystyle h^2}}\right]\\ & =& 0\end{array}$

Hence, by the definition of derivative the function $f$ is differentiable at $k=0$.

Tho objective is to use the Maclaurin series for the function $f\left(x\right)=\cos x$to find the Maclaurin series for the function $\frac{1-\cos x}{x}$.
The Maclaurin series for the function $f\left(x\right)=\cos x$is $\cos x=\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$.
Therefore, the Maclaurin series for the function $\frac{1-\cos x}{x}$is $\frac{1-\cos x}{x}=\frac{1}{x}\left[1-\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}\right]$.