Here is a theorem which I came across in Principles of Mathematical analysis by Walter Rudin
Theorem 7.17$\space\space$ Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a,b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b].$ If $\{f_n'\}$ converges uniformly on $[a,b]$, then ${f_n}$ converges uniformly on $[a,b]$, to a function $f$, and $$f'(x)=\lim_{n\to\infty}f'_n(x)\quad\quad\quad\quad(a\le x\le b)$$
In the proof of this theorem he proves the uniform convergence of ${f_n}$ and defines the limit function as equal to $f$. Further he says that if we fix a point $x$ then we can define $$\phi_n(t)=\frac{f_n(t)-f_n(x)}{t-x},\quad\quad\quad \phi(t)=\frac{f(t)-f(x)}{t-x}$$ for $a\le t\le b,\space t\neq x$. It is then proven that $\phi_n(t)$ converges uniformly for $t\neq x$ using the Cauchy criterion of uniform convergence which means that the proof of $\phi_n(t)$ is uniformly convergent has been established but the limit function is yet unknown. He then writes that since $\{f_n\}$ converges to $f$ therefore we can conclude that $$\lim_{n\to \infty}\phi_n(t)=\phi(t)$$ It is this part of the proof which I don't understand. If we are claiming that $\phi_n(t)$ converges uniformly to $\phi(t)$ then it should be possible to show that for an $\varepsilon \gt 0$ there exists an $N$ such that $\quad sup|\phi_n(t)-\phi(t)|\lt\varepsilon\quad$ for all $n\ge N$ if $t\in[a,b],$ and $t\neq x$. The way I started to prove it is
$$|\phi_n(t)-\phi(t)|=\left|\frac{f_n(t)-f_n(x)}{t-x}-\frac{f(t)-f(x)}{t-x}\right|\le \frac{1}{|t-x|} \Bigl[|f_n(t)-f(t)|+|f_n(x)-f(x)|\Bigr]$$ If I resolve the above I will be able to show the existence of an $N$ such that $$|\phi_n(t)-\phi(t)| \lt \frac{\varepsilon}{|t-x|}$$ for all $n\ge N$. So, how do I get rid of $|t-x|$ in the above since I can't assume that $f$ is differentiable (that is what we have to prove in this theorem) so mean value theorem is out of question?
Another doubt in the same proof has already been asked here: Baby Rudin Theorem 7.17 where the whole proof has been quoted from the book. Since I don't have enough reputation yet so I couldn't ask for clarification of the proof in the same thread itself and hence have to post it as a separate question.
As I mentioned in my comment, one could avoid an explicit usage of the Cauchy criterion by intertwining it into a complicated estimate (i.e basically repeat proofs of theorems already proven). Just to be clear: I think Rudin's proof is going to be much simpler than what I present below, but in case you're a brave soul, here goes.
Let $\psi$ be the limit of $f_n'$. I'm going to assume we already know $f_n$ converges pointwise to some function $f$. Now, fix an $x\in [a,b]$. Now, for any $t\neq x$, and any $n,m\in\Bbb{N}$, we use a 3-fold triangle inequality to get \begin{align} \left|\frac{f(t)-f(x)}{t-x}-\psi(x)\right|&\leq \left|\frac{f(t)-f(x)}{t-x}-\frac{f_n(t)-f_n(x)}{t-x}\right|\\ &+\left|\frac{(f_n-f_m)(t)-(f_n-f_m)(x)}{t-x}\right|\\ &+\left|\frac{f_m(t)-f_m(x)}{t-x}-\psi(x)\right| \end{align} Using the mean-value theorem for $f_n-f_m$, we can bound the second term by $\|f_n'-f_m'\|_{\infty}$ (i.e the supremum norm). Hence, we get \begin{align} \left|\frac{f(t)-f(x)}{t-x}-\psi(x)\right|&\leq \left|\frac{f(t)-f(x)}{t-x}-\frac{f_n(t)-f_n(x)}{t-x}\right|\\ &+\|f_n'-f_m'\|_{\infty}\\ &+\left|\frac{f_m(t)-f_m(x)}{t-x}-\psi(x)\right| \end{align} Now, we take the limit as $n\to \infty$ on both sides. On the LHS, nothing happens, while on the RHS, the first term vanishes since $f_n\to f$ pointwise by definition. The second term becomes $\|\psi-f_m'\|_{\infty}$ (because norms are continuous so we can pull the limit inside. Note that we're using the fact $f_n'\to \psi$ uniformly here). Hence, \begin{align} \left|\frac{f(t)-f(x)}{t-x}-\psi(x)\right| & \leq 0 + \|\psi-f_m'\|_{\infty}+ \left|\frac{f_m(t)-f_m(x)}{t-x}-\psi(x)\right| \end{align} Now, we take $\limsup\limits_{t\to x}$ on both sides (we take $\limsup$ because the limit superior of any function always exists, while the limit a-priori need not exist. Also, recall that limit exists if and only if $\limsup$ and $\liminf$ are equal, in which case all three are equal). This gives us \begin{align} \limsup_{t\to x}\left|\frac{f(t)-f(x)}{t-x}-\psi(x)\right| & \leq \|\psi-f_m'\|_{\infty}+\limsup_{t\to x}\left|\frac{f_m(t)-f_m(x)}{t-x}-\psi(x)\right|\\ &=\|\psi-f_m'\|_{\infty}+|f_m'(x)-\psi(x)|\\ &\leq 2\|f_m'-\psi\|_{\infty}. \end{align} Finally, we can let $m\to \infty$, so that the RHS vanishes. Therefore, we have proven that \begin{align} \limsup_{t\to x}\left|\frac{f(t)-f(x)}{t-x}-\psi(x)\right|&\leq 0. \end{align} This means the $\limsup$ is actually equal to $0$, and thus the $\lim\limits_{t\to x}$ also exists and equals zero. This is exactly what it means to show $f$ is differentiable at $x$ with $f'(x)=\psi(x)=\lim\limits_{n\to \infty}f_n'(x)$.
The usefulness of $\limsup$ is that it saves us from bringing in $\epsilon$'s and $N$'s and $\delta$'s.