# Exercise 4.1.14 Introduction to Real Analysis by Jiri Lebl

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.

$$|a+b(x-c) - f(x)|$$ 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 $$\delta$$.
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.