For n ∈ $\mathbb{N}$, let $f_n : [−1, 1] \rightarrow \mathbb{R}, f_n(x) = n(\exp(x/n)−1)$.
a) Show that the sequence of functions $(f_n)$ converges pointwise on $[−1, 1]$, and determine the limit function $f$.
b) Show that the family ${f_n : n ∈ \mathbb{N}} ⊂ C([−1, 1], \mathbb{R})$ is pointwise bounded and equicontinuous on $[−1, 1]$.
I am mostly having trouble with the Taylor expansions when trying to prove the inequalities (see below). I've done the Taylor expansions wrt $x$ close to zero, however $x$ lies in the interval $[-1,1]$. For clarity I've included my attempt on the problem, with exclamation marks on top of equalities which I am not sure if they are correct.
My Attempt
a) Def: $(f_n) \underset{n \rightarrow \infty}{\rightarrow} f$ pointwise $\iff \forall \epsilon > 0 \hspace{2mm} \forall x \in [0,1] \hspace{2mm} \exists N \in \mathbb{N} \hspace{1mm}$ s.t $d(f_n(x),f(x)) < \epsilon \hspace{2mm} \forall n > N$. In other words $(f_n(x)) \underset{n \rightarrow \infty}{\rightarrow} f(x)$ for each $x$ seperately.
$ \lim{n \to \infty} \hspace{1mm} f_n(x) = \lim{n \to \infty} \hspace{2mm} n(\exp(x/n)−1) \overset{!}{\approx} \lim{n \to \infty} \hspace{2mm} n \Big( 1 + \frac{x}{n} + \mathcal{O}(\frac{1}{n^2}) -1 \Big) = x \Rightarrow f(x) = x$
I do the Taylor expansion of $\exp(x/n)$ near $x=0$. Considering that $x$ lies in the interval $[-1,1]$, is this motivated? If I do the expansion around an arbitrary point $x=a$ I instead end up with $\exp(a/n) (x-a)$ and I do not know how to find a limit function $f(x)$ from this.
b) Def: $(f_n) \subset Y^X$ pointwise bounded if $(f_n(x)) \subset Y$ bounded in $(Y,d)$.
Again doing a Taylor expansion near $x=0$, I end up with $f_n(x) \overset{!}{\approx} n \Big( 1 + \frac{1}{n} x + \mathcal{O}(\frac{1}{n^2}) -1 \Big) = x + \mathcal{O}(\frac{1}{n}) < \infty \hspace{2mm} \forall n>0 \hspace{2mm} \forall x \in [-1,1]$.
Def: $(f_n)$ equicontinuous if $\forall \epsilon >0 \hspace{2mm} \forall x_0 \in [-1,1] $ there exists an open nbh $U$ of $x_0$ s.t $\forall x \in U \hspace{2mm} \forall f_n \in (f_n)$ we have $d(f_n(x),f_n(x_0)) < \epsilon$
To show this, define the open nbh of $U$ as $x = x_0 + \delta$ and prove $d(f_n(x_0 + \delta),f_n(x_0)) < \epsilon$. This one I felt was quite straightforward as one could do the Taylor expansion for $\delta$ close to zero, which is motivated by us picking a small nbh $U$ of $x_0$.
To prove pointwise convergence you could use L'Hospitals rule while fixing x. $\newcommand{\nlim}{\lim\limits_{n \to \infty}}$$\nlim n\exp\left(\frac{x}{n}-1\right)=\nlim \frac{\exp\left(\frac{x}{n}-1\right)}{\frac{1}{n}}=\nlim \frac{\frac{-x}{n^2}\exp\left(\frac{x}{n}-1\right)}{\frac{-1}{n^2}}=\nlim x\exp\left(\frac{x}{n}-1\right)=\frac{x}{e}$
The pointwise boundedness follows from the pointwise convergence.
Coming to the equicontinuity, I don't know if you mean $f_n$ or the pointwise limit of $f_n$, but it is true for both. Every $f_n$ is obviously equicontinuous because $f_n$ is continuous and $[0, 1]$ is compact. If you define $f(x):= \nlim f_n(x)= \frac{x}{e} \quad\forall x \in [0,1]$, then $f$ is also equicontinuous, because $f$ is continuous and $[0,1]$ is compact.