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Suppose $f'$ is continuous on $[a,b]$ and $\epsilon>0$. Prove that there exists $\delta>0$ such that $$\left|\frac{f(t)-f(x)}{t-x}-f'(x)\right|<\epsilon$$ whenever $0<|t-x|<\delta$, with $x$ and $t$ in $[a,b$].

Since $f'$ is continuous on a compact space $[a,b]$, it is uniformly continuous on $[a,b]$. So for any $\epsilon >0$, there exists $\delta$ such that if $|t-x|<\delta$, then $|f'(t)-f'(x)|<\epsilon$.

Now for any $x$ and any $\epsilon>0$, there exists $\delta_x$ such that if $|t-x|<\delta_x$, then $\left|\dfrac{f(t)-f(x)}{t-x}-f'(x)\right|<\epsilon$. But this $\delta_x$ depends on $x$, and so I'm not able to conclude.

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Did you try using Lagrange's theorem, say? That is, to write $$\frac{f(t)-f(x)}{x-t}$$ in a more convenient form. –  Pedro Tamaroff Jun 21 '13 at 3:33
    
@PeterTamaroff I've never heard of it. –  PJ Miller Jun 21 '13 at 3:34
    
Do you know Rolle's theorem? –  Pedro Tamaroff Jun 21 '13 at 3:35
    
@PeterTamaroff Yes, as well as mean-value theorem. –  PJ Miller Jun 21 '13 at 3:37
2  
I see. So $\frac{f(t)-f(x)}{t-x}$ must equal $f'(c)$ for some $c$ between $t$ and $x$. Combine that with the first paragraph in my post, and that should finish it. –  PJ Miller Jun 21 '13 at 3:42

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