I have a question about Theorem 7.17 in the book *Principles of Mathematical Analysis* by Walter Rudin, 3rd edition:

> 
> Suppose $\left\{ f\_n \right\}$ is a sequence of functions, differentiable on $[a, b]$ and such that $\left\{ f\_n \left( x\_0 \right) \right\}$ converges for some point $x\_0$ on $[a, b]$. If $\left\{ f\_n^\prime \right\}$ converges uniformly on $[a, b]$, then $\left\{ f\_n \right\}$ converges uniformly on $[a, b]$, to a function $f$, and
> $$\tag{27} f^\prime(x) = \lim\_{n \to \infty } f\_n^\prime(x) \qquad \qquad (a \leq x \leq b). $$
> 
> 
> 

Rudin's proof is as follows:

> 
> Let $\varepsilon > 0$ be given. Choose $N$ such that $n \geq N$, $m \geq N$, implies
> $$ \tag{28} \left\lvert f\_n \left( x\_0 \right) - f\_m \left( x\_0 \right) \right\rvert < \frac{\varepsilon}{2} $$
> and
> $$ \tag{29} \left\lvert f\_n^\prime(t) - f\_m^\prime(t) \right\rvert < \frac{\varepsilon}{2(b-a)} \qquad \qquad (a \leq t \leq b). $$
> 
> 
> 

> 
> If we apply the mean value theorem 5.19 to the function $f\_n - f\_m$, (29) shows that
> $$ \tag{30} \left\lvert f\_n (x) - f\_m (x) - f\_n (t) + f\_m (t) \right\rvert \leq \frac{ \lvert x-t \rvert \varepsilon }{ 2(b-a) } \leq \frac{ \varepsilon }{ 2 } $$
> for any $x$ and $t$ on $[a, b]$, if $n \geq N$, $m \geq N$. The inequality
> $$ \left\lvert f\_n (x) - f\_m (x) \right\rvert \leq \left\lvert f\_n (x) - f\_m (x) - f\_n \left( x\_0 \right) + f\_m \left( x\_0 \right) \right\rvert + \left\lvert f\_n \left( x\_0 \right) - f\_m \left( x\_0 \right) \right\rvert $$
> implies, by (28) and (30), that
> $$ \left\lvert f\_n(x) - f\_m(x) \right\rvert < \varepsilon \qquad \qquad (a \leq x \leq b, n \geq N, m \geq N), $$
> so that $\left\{ f\_n \right\}$ converges uniformly on $[a, b]$. Let
> $$ f(x) = \lim\_{n \to \infty} f\_n (x) \qquad (a \leq x \leq b). $$
> 
> 
> 

> 
> Let us now fix a point $x$ on $[a, b]$ and define
> $$ \tag{31} \phi\_n(t) = \frac{ f\_n(t) - f\_n(x) }{ t-x}, \qquad \phi(t) = \frac{ f(t) - f(x) }{ t-x} $$
> for $a \leq t \leq b$, $t \neq x$. Then
> $$ \tag{32} \lim\_{ t \to x } \phi\_n (t) = f\_n^\prime(x) \qquad (n = 1, 2, 3, \ldots). $$
> The first inequality in (30) shows that
> $$ \left\lvert \phi\_n(t) - \phi\_m(t) \right\rvert \leq \frac{ \varepsilon}{2(b-a) } \qquad ( n \geq N, m \geq N), $$
> so that $\left\{ \phi\_n \right\}$ converges uniformly, for $t \neq x$. Since $\left\{ f\_n \right\}$ converges to $f$, we conclude from (31) that
> $$\tag{33} \lim\_{n \to \infty } \phi\_n (t) = \phi(t) $$
> uniformly for $a \leq t \leq b$, $t \neq x$.
> 
> 
> 

> 
> If we now apply Theorem 7.11 to $\left\{ \phi\_n \right\}$, (32) and (33) show that
> $$ \lim\_{t \to x} \phi(t) = \lim\_{n \to \infty} f\_n^\prime(x); $$
> and this is (27), by the definition of $\phi(t)$.
> 
> 
> 

How is the claim on line 33 proven? From my understanding, Rudin has shown only that ${\phi\_n(t)}$ converges uniformly by the Cauchy criterion for uniform convergence, but it is unclear to me as to why it converges in particular to $\phi(t)$.