A doubt about the proof about the differentiability of the limits of sequences of functions

Question

I'm reading a theorem about the differentiability of the limits of sequences of functions:

and its proof:

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1. The Mean Value Theorem is as follows:

Should it be $$\left|f_{n}(x)-f_{n}(a)-f_{n}^{\prime}(a)(x-a)\right| \leq \sup _{0<t<1}\left|f_{n}^{\prime}(a+t(x-a))-f_{n}^{\prime}(a)\right|\color{blue}{|1-0|}$$ rather than $$\left|f_{n}(x)-f_{n}(a)-f_{n}^{\prime}(a)(x-a)\right| \leq \sup _{0<t<1}\left|f_{n}^{\prime}(a+t(x-a))-f_{n}^{\prime}(a)\right|\color{blue}{|x-a|}$$ ?

2. In a similar manner, should it be $$\left|f_{n}(x)-f(x)-\left(f_{n}(a)-f(a)\right)\right| \le \color{blue}{|1-0|} \sup _{0<t<1}\left|f_{n}^{\prime}(a+t(x-a))-f^{\prime}(a+t(x-a))\right|$$ rather than $$\left|f_{n}(x)-f(x)-\left(f_{n}(a)-f(a)\right)\right| \le \color{blue}{r} \sup _{0<t<1}\left|f_{n}^{\prime}(a+t(x-a))-f^{\prime}(a+t(x-a))\right|$$?

3. I would like to confirm my understanding about how 2.1 implies $$f(x)-f(a)-g(a)(x-a)=o(|x-a|) \quad(x \rightarrow a)$$

Let $M_x = \sup _{0<t<1}|g(a+t(x-a))-g(a)|$. Then $$\begin{aligned} \lim_{x \to a} M_x &= \lim_{x \to a} \sup _{0<t<1} |g(a+t(x-a))-g(a)| \\&= \sup _{0<t<1} \lim_{x \to a} \ |g(a+t(x-a))-g(a)|\\ &= \sup _{0<t<1} 0 \\ &=0\end{aligned}$$

It follows from 2.1 that $$\frac{|f(x)-f(a)-g(a)(x-a)|}{|x-a|} \le \sup _{0<t<1}|g(a+t(x-a))-g(a)| = M_x$$

and thus $$\lim_{x \to a}\frac{|f(x)-f(a)-g(a)(x-a)|}{|x-a|} \le \lim_{x \to a} M_x = 0$$

and consequently $$\lim_{x \to a}\frac{|f(x)-f(a)-g(a)(x-a)|}{|x-a|} = \lim_{x \to a} \left | \frac{f(x)-f(a)-g(a)(x-a)}{x-a} \right | = 0$$

and consequently $$\lim_{x \to a} \frac{f(x)-f(a)-g(a)(x-a)}{x-a} = 0$$, which means $g(a) =f'(a)$

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My question:

Could you please confirm of my above understanding is correct? Thank you so much for your help!

Answer 1

Define $\phi:[0,1]\rightarrow E,t\mapsto f_n(a+t(x-a))-tf_n'(a)(x-a)$ then $$|\frac{\phi(1)-\phi(0)}{1-0}|=|f_n(x)-f_n'(a)(x-a)-f_n(a)|\leq\sup_{0<t<1}|\phi'(t)|$$ and using the chain rule differentiating with respect to $t$ You get $$|\phi'(t)|=|f_n'(a+t(x-a))(x-a)-f_n'(a)(x-a)|\leq |f_n'(a+t(x-a)))-f_n'(a)||x-a|.$$ Similar with the other function. The factor $x-a$ is the differentiation of the inner function.