# Best approximation

• Nov 1st 2009, 07:27 PM
cgiulz
Best approximation
Prove:

If $\displaystyle f(x)$ is defined for $\displaystyle x \approx a,$ and there is a number $\displaystyle k$ such that

$\displaystyle f(x) = f(a) + k(x - a) + e(x),$ where $\displaystyle \displaystyle\lim_{x \to a}\frac{e(x)}{x - a} = 0,$

then $\displaystyle f(x)$ is differentiable at $\displaystyle a,$ and $\displaystyle k = f'(a).$
• Nov 2nd 2009, 03:14 AM
HallsofIvy
Quote:

Originally Posted by cgiulz
Prove:

If $\displaystyle f(x)$ is defined for $\displaystyle x \approx a,$ and there is a number $\displaystyle k$ such that

$\displaystyle f(x) = f(a) + k(x - a) + e(x),$ where $\displaystyle \displaystyle\lim_{x \to a}\frac{e(x)}{x - a} = 0,$

then $\displaystyle f(x)$ is differentiable at $\displaystyle a,$ and $\displaystyle k = f'(a).$

My first thought was "what is your definition of 'differentiable'?". Can we take it that this is in R and we are using the Calculus I definition of the derivative: $\displaystyle f'(a)= \lim_{h\to 0}\frac{f(a+h)- f(a)}{h}$? If so, just put x= a+h in your formula for f(x) and put that into the difference quotient.