ziphumulegn

2022-07-09

Since we know that in a good linear approximation, $L\left(x\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right).$. But what if ${f}^{\prime }\left(a\right)$ does not exist? How to prove that if a function has a good linear approximation, then it must be differentiable?

Dobermann82

Expert

One definition of having a good linear approximation for $f$ at $a$ is
$f\left(x\right)=f\left(a\right)+c\left(x-a\right)+ϵ\left(x-a\right)$
where $ϵ\left(x-a\right)$, the error term, is a function that satisfies the following
$\underset{x\to a}{lim}\frac{ϵ\left(x-a\right)}{x-a}=0$
Then, that $f$ is differentiable basically follows immediately, by subtracting $f\left(a\right)$, dividing by x−a and taking the limit as $x-a$ showing that the function is differentiable at $a$, with value $c$.
$f\left(x\right)=f\left(a\right)+c\left(x-a\right)+ϵ\left(x-a\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(x\right)-f\left(a\right)=c\left(x-a\right)+ϵ\left(x-a\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{f\left(x\right)-f\left(a\right)}{x-a}=c+\frac{ϵ\left(x-a\right)}{x-a}$
and thus taking the limit
$\underset{x\to a}{lim}\frac{f\left(x\right)-f\left(a\right)}{x-a}=c+\underset{x\to a}{lim}\frac{ϵ\left(x-a\right)}{x-a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{f}^{\prime }\left(a\right)=c$

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