$f_{n}$ uniformly converges to $f$ and $|f'_{n} (x)| ≤ C $, is $f$ necessarily differentiable? Assume $f(x)$ is the uniform limit of real differentiable functions $f_{n}(x)$ on $[−1,1]$. Assume that $|f'_{n} (x)| ≤ C $ for some $C$ independent of $n$ and $x ∈ [−1, 1]$. Is the function $f(x)$ necessarily differentiable?

No. Choose $f_n(x) := \sqrt{x^2 + \frac{1}{n}}$. Then, for all $x \in [-1, 1]$:
$$
\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert = \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \leq \lvert x \rvert + \frac{1}{\sqrt{n}} - \lvert x \rvert = \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longrightarrow}0
$$
Therefore
$$
\sup_{x \in [-1, 1]}\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert \leq \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longrightarrow}0
$$
and thus uniform convergence. The limit, i.e. the absolute value is not differentiable in $0$. Furthermore, for all $n \in \mathbb{N}$, $x \in [-1, 1]$:
$$
\lvert f_n'(x) \rvert = \left \lvert \frac{x}{\sqrt{x^2 + \frac{1}{n}}} \right \rvert = \frac{\sqrt{x^2}}{\sqrt{x^2 + \frac{1}{n}}} \leq \frac{\sqrt{x^2 + \frac{1}{n}}}{\sqrt{x^2 + \frac{1}{n}}} = 1
$$
