How can i prove this question without using The Mean Value Theorem? Question:
Suppose $f'$ is continuous on $[a,b]$ and $\epsilon\gt0$.Prove that there exist $\delta\gt0$ such that $|\frac{f(t)-f(x)}{t-x}-f'(x)|\lt\epsilon$ whenever $0\lt|t-x|\lt\delta$,with $x$ and $t$ in $[a,b]$.
My question is how can I prove it without using The Mean Value Theorem.
 A: Suppose it were false. Then there exists $r>0$ and there exist sequences $(t_n)_n$ and $(x_n)_n$ in $[a,b]$ with $t_n\ne x_n$ and with
$t_n,x_n$ both converging to some $y\in [a,b],$ and such that $$(*)\quad \left|\frac {f(t_n)-f(x_n)}{t_n-x_n}-f'(x_n)\right|>r$$ for every $n.$
Now $$\frac {f(t_n)-f(x_n)}{t_n-x_n}=\frac {1}{t_n-x_n}\int_{x_n}^{t_n}f'(u)du$$ is bounded above by $\max_{u\in [x_n,t_n]}f'(u)$ and bounded below by $\min_{u\in [x_n,t_n]}f'(u)$. So by $(*)$ there exists $u_n\in [x_n,t_n]$ with $$|f'(u_n)-f'(x_n)|>r.$$
But  $u_n$ and $x_n$ both converge to $y$ so $f'$ is not continuous at $y,$ a contradiction.
A: I had the same idea as Adayah's suggestion in his comment (which I hadn't read before) of DanielWainfleet's answer.
Since $f'$ is continuous on $[a,b]$, it is uniformly continuous, hence there exists $\delta>0$ such that forall $u,x\in[a,b]$:
$$|u-x|<\delta\Longrightarrow|f'(u)-f'(x)|<\epsilon.$$
Then, forall $x,t\in[a,b]$ such that $0<|t-x|<\delta$:
$$\left|\frac{f(t)-f(x)}{t-x}-f'(x)\right|=\left|\frac1{t-x}\int_x^t(f'(u)-f'(x))\,\mathrm du\right|<\epsilon.$$
