Q. 83 - Calculus

Question

Let, $f\left(x\right)=\left\{\begin{array}{l}\frac{\sin x}{x}, if x\ne 0\\ 0 , if x=0\end{array}\right.$

(a) Use the definition of the derivative to prove that $f$ is differentiable at $0$

(b) Use the Maclaurin series for $\sin x$ to find a Maclaurin
series for $f$

Step-by-Step Solution

Answer

(a) $f$ is differentiable at $0$ is proved

(b) The required Maclaurin series is $\frac{1}{x}\left(\sum _{k=1}^x\frac{(-1{)}^d}{(2k+1)!}{x}^{2a+1}\right)$

Step 1 Given Information

Consider the function, 

$f\left(x\right)=\left\{\begin{array}{l}\frac{\sin x}{x}, if x\ne 0\\ 0 , if x=0\end{array}\right.$

Part (a) Step 1 Derivative test

Derivative test states that,
If $f$ is a function defined from $(a,b)\to R$ and $k\in (a,b)$, then $f$ is differentiable at $k$ if the limit $\underset{h\to \infty }{\lim }\frac{f(k+h)-f\left(h\right)}{h}$ 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 $k=0$.

Part (a) Step 2 Proof

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

Consider the following expression and solve as.
$\begin{array}{rcl}\underset{h\to \infty }{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}& =& \underset{h\to \infty }{\lim }\left[\frac{\frac{\sin h}{h}-1}{h}\right]\\ & =& 0\end{array}$
Hence, the function $f$ is differentiable at $k=0$.

Part (b) Step 1 Calculation

The objective is to use the Maclaurin series for the function $f\left(x\right)=\sin x$ to find the Maclaurin series for the function $\frac{\sin x}{x}$.
The Maclaurin series for the function $f\left(x\right)=\sin x$ is $\sum _{k=1}^2\frac{(-1{)}^k}{(2k+1)!}{x}^{2k+1}$.
Therefore, the Maclaurin series for the function $\frac{\sin x}{x}$is, $\frac{1}{x}\left(\sum _{k=1}^x\frac{(-1{)}^d}{(2k+1)!}{x}^{2a+1}\right)$.