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Suppose $f : I \to \mathbb{R}$ is diffrentiable at $c \in I$. Prove there exist numbers $a$ and $b$ with the property that for every $\epsilon >0$, there is a $\delta >0$, such that $|a + b (x-c) - f(x)| \le \epsilon|x-c|$, whenever $x \in I$ and $|x-c| < \delta$. In other words, show that there exists a function $g: I \to \mathbb{R}$ such that $\lim_{x\to c} g(x) = 0 $ and $| a + b(x-c) - f(x)| \le |x -c | g(x)$.

I don't know how to start this question. I think that I should use the hypothesis that $f$ is differentiable at $c$, but I don't know how to use this.

I appreciate if you give some help.

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$|a+b(x-c) - f(x)|$ is the distance between a line $a + b(x-c)$ and your function $f(x)$ when you plug in $x$. If you want this to be small for $x$ near $c$, it might make sense to use the line tangent to $f$ at $c$.

That is, use the line that intersects the point $(c, f(c))$ and has slope $f'(c)$. Taking $a=f(c)$ and $b=f'(c)$ does the trick.

Now, given $\epsilon > 0$, you want to find $\delta>0$ such that for any $x$ satisfying $|x-c|< \delta$, you also have $$\frac{|f(c) + f'(c) (x-c) - f(x)|}{|x-c|} < \epsilon.$$

Indeed, note $$\frac{|f(c) + f'(c) (x-c) - f(x)|}{|x-c|} = \left|f'(c) - \frac{f(x) - f(c)}{x-c}\right|$$ and use the limit definition of a derivative ($f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x-c}$) to find the appropriate $\delta$.

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Let $g(x)=\left|\dfrac{f(x)-f(c)}{x-c}-f'(c)\right|$ for $x\ne c$ and $g(c)=0$, this will do the job.

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