Continuity, Differentiability, and \mathbb{N}: Showing an Inequality

[.....]

This isn't homework per se... It's a question from a book I'm self-studying from.

If $$f$$ is continuous on $$[a,b]$$ and differentiable at a point $$c \in [a,b]$$, show that, for some pair $$m,n \in \mathbb{N}$$,

$$\left | \frac{f(x)-f(c)}{x-c}\right | \leq n$$ whenever $$0 \leq |x-c| \leq \frac{1}{m}$$

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Since it's differentiable at c I know $$\lim_{x \to c}\frac{f(x)-f(c)}{x-c}$$ exists...

And it's continuous so I know for whatever $$\epsilon > 0$$ I pick there's a $$\delta > 0$$ so that $$|f(x)-f(c)| < \epsilon$$ when $$|x-c| < \delta$$

So I guess 1/m could be delta? And then... ? Not sure how to round this one off

 

[phsopher]

Couldn't you just conclude that:

$$\left | \frac{f(x)-f(c)}{x-c}\right | \leq n \Leftrightarrow | f(x)-f(c) | \leq n | x - c | \leq \frac{n}{m}$$

and that you can always find m and n such that this holds as long as $$| f(x)-f(c) |$$ is finite which is ensured by differentiability?

 

[Shooting Star]

So I guess 1/m could be delta? And then... ?

Pick m as the integral part of 1/δ so that 1/δ = m+f. Then δ = 1/(m+f) < 1/m.