Quadratic approximation of $\frac{sin(x)}x$ at $x=0$

Question

Write down the quadratic $f(0)+f'(0)\Delta x+\frac12f''(0)(\Delta x)^2$ for $f(x)=\frac{sin(x)}x$.

$f(0)$, $f'(0)$ and $f''(0)$ are not defined. How could we produce a quadratic approximation at $x=0$?

Answer 1

I think you are asking about the correctness of the question, rather than the expected answer.

You are absolutely right to wonder about this. $f(x)$ is not defined at $x=0$, so what can $f(0), f'(0), f''(0)$ mean? The question is badly posed.

But in truth it is obvious what the question intends: it is about the function $f$ defined by

$$ f(x) = \begin{cases} 1, & \text{if $x = 0$} \\ \dfrac{\sin x}{x}, & \text{if $x \ne 0$} \end{cases} $$

And this function is infinitely differentiable at $x=0$.

Answer 2

If you call

$$ f(x) = \frac{\sin x}{x} $$

Use the limit

\begin{eqnarray} \lim_{x \to 0} f(x) &=& 1 \\ \lim_{x \to 0} f'(x) &=& 0 \\ \lim_{x \to 0} f''(x) &=& -1/3 \\ \end{eqnarray}

So that you have

$$ f(x) = 1 - x^2/6 + O(x^3) $$

Answer 3

You could recall that

$$\sin(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

which is the famous expansion for $\sin(x)$. Then simply divide both sides by $x$.