Pointwise convergent of the sequence of functions Prove that if $\{f_n : \mathbb R \rightarrow \mathbb R\}$ is a sequence of continuously differentiable functions such that the sequence of derivatives $\{f'_n : \mathbb R \rightarrow \mathbb R\}$ is uniformly convergent and the sequence $\{f_n (0)\}$ is also convergent, then $\{f_n : \mathbb R \rightarrow \mathbb R\}$ is pointwise convergent. Is the assumption that the sequence $\{f_n (0)\}$ converges necessary?


My attempt: Suppose $f'_n \rightarrow g$ uniformly and $f_n (0) \rightarrow a$. Then since $f'_n$ are integrable, $\int_0^x f'_n \rightarrow \int_0^x g$, i.e. $\lim_{n \rightarrow \infty} \int_0^x f'_n = \lim_{n \rightarrow \infty} (f_n(x)-f_n(0))=\int_0^x g$. Is this the right approach, cause I'm stuck here, if so, how to proceed?
$\newcommand{\dif}{\,\text{d}}$
By the fundamental theorem of calculus we have $f_n(x) = f_n(0) + \int_0^x f'_n(t) \dif t$. You know $f_n(0)$ converges to something and you know $\int_0^x f'_n(t) \dif t$ converges to something. Putting this together shows $f_n(x)$ converges to something.
$\newcommand{\dif}{\,\text{d}}$
By the fundamental theorem of calculus we have $f_n(x) = f_n(0) + \int_0^x f'_n(t) \dif t$. You know $f_n(0)$ converges to something and you know $\int_0^x f'_n(t) \dif t$ converges to something. Putting this together shows $f_n(x)$ converges to something.