ziphumulegn

Answered

2022-07-09

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

Answer & Explanation

Dobermann82

Expert

2022-07-10Added 15 answers

One definition of having a good linear approximation for $f$ at $a$ is

$f(x)=f(a)+c(x-a)+\u03f5(x-a)$

where $\u03f5(x-a)$, the error term, is a function that satisfies the following

$\underset{x\to a}{lim}\frac{\u03f5(x-a)}{x-a}=0$

Then, that $f$ is differentiable basically follows immediately, by subtracting $f(a)$, dividing by x−a and taking the limit as $x-a$ showing that the function is differentiable at $a$, with value $c$.

$f(x)=f(a)+c(x-a)+\u03f5(x-a)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(x)-f(a)=c(x-a)+\u03f5(x-a)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{f(x)-f(a)}{x-a}=c+\frac{\u03f5(x-a)}{x-a}$

and thus taking the limit

$\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=c+\underset{x\to a}{lim}\frac{\u03f5(x-a)}{x-a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{f}^{\prime}(a)=c$

$f(x)=f(a)+c(x-a)+\u03f5(x-a)$

where $\u03f5(x-a)$, the error term, is a function that satisfies the following

$\underset{x\to a}{lim}\frac{\u03f5(x-a)}{x-a}=0$

Then, that $f$ is differentiable basically follows immediately, by subtracting $f(a)$, dividing by x−a and taking the limit as $x-a$ showing that the function is differentiable at $a$, with value $c$.

$f(x)=f(a)+c(x-a)+\u03f5(x-a)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(x)-f(a)=c(x-a)+\u03f5(x-a)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{f(x)-f(a)}{x-a}=c+\frac{\u03f5(x-a)}{x-a}$

and thus taking the limit

$\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=c+\underset{x\to a}{lim}\frac{\u03f5(x-a)}{x-a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{f}^{\prime}(a)=c$

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