Check $f_n(x) := n\ln(1 + \frac{2^x}{n})\ x \in (0, 1)$ for pointwise and uniform convergence.
I am able to show that $\ f_n(x)$ converges pointwise to $2^x$ (just apply exponentiation). However, it is hard to tell something about uniform convergence.
For example, if I take $2^x$ as $f(x)$ and want to show that $f_n(x)$ uniformly converges to $f(x)$, I have to find or at least estimate $\sup \limits_{x \in(0, 1)} |n\ln(1 + \frac{2^x}{n}) - 2^x|$, and this is the point where I got stuck.
Since $[0,1]$ is compact, and you have pointwise convergence already and $f_n(x)\nearrow$ at $n$ fixed, and $2^x$ continuous, then second Dini's theorem applies here.
wikipedia: Deuxième théorème de Dini
[I don't know why, but there is only the first Dini's theorem on English language wiki].
You should not be afraid that the $f_n$ are only defined only $(0,1)$ since they are trivially extendable continuously on $[0,1]$.
Note that you can arrive at this result manually since
$n\ln\left(1+\dfrac {2^x}n\right)-2^x=n\left(\dfrac{2^x}n-\dfrac{4^x}{2n^2}+o(\dfrac{4^x}{2n^2})\right)-2^x=-\dfrac{4^x}{2n}+o(\dfrac{4^x}{2n})$
And since $|4^x|\le|4^1|\le 4$ is bounded, all this is $\bigg|n\ln\left(1+\dfrac {2^x}n\right)-2^x\bigg|<\dfrac kn$ for some $k$ when $n\ge n_0\gg 1$.
And you get your uniform convergence.
EDIT:
The Taylor expansion I think is straightforward, so I guess your concern is why
$\bigg|-\dfrac{4^x}{2n}+o(\dfrac{4^x}{2n})\bigg|<\dfrac kn$ ?
First thing is that we work for $x\in[0,1]$ so $1\le 4^x\le 4$, this is essential for uniform convergence because this is here that we get rid of the dependency in $x$.
So we get $0\le\dfrac{4^x}{2n}\le \dfrac 4{2n}\le \dfrac 2n$
Secondly by definition of $o$, for any $\varepsilon$ and $n$ big enough, or let say $\forall n\ge n_0\gg 1$ we have
$\bigg|o(\dfrac{4^x}{2n})\bigg|\le\bigg|o(\dfrac{2}{n})\bigg|\le \dfrac{2\varepsilon}{n}\quad$ and the choice of $n_0$ is independant of $x$.
Finally $\bigg|f_n(x)-2^x\bigg|\le \dfrac{2+\varepsilon}{n}\le \dfrac 3n$ for $\varepsilon\le 1$ for instance.
And we have an upper bound that depends only on $n$ and goes to $0$ for an $n_0$ that does not depend on $x$, this is the definition of uniform convergence.