With l'hopital's rule I can show that $f(x)=2^x\log{(1+2^{-x})}$ goes to $1$ as $x$ goes to $\infty$.
What I intuitively don't get though is why apparently $2^x$ and $\log{(1+2^{-x})}$ exactly balance each other such that the limit to $\infty$ is $1$.
For example, if the base of the first term would be slightly higher, then the limit would go to $\infty$.
Any hints for the intuition of this result?
If you let $2^x=u$, you're really asking about $u\ln(1+\frac1u)$. Notice then that
$$\lim_{u\to\infty}u\ln\left(1+\frac1u\right)=\lim_{u\to\infty}\ln\left(1+\frac1u\right)^u=\ln(L)$$
Indeed, it is well known that
$$L=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$
which happens to be one of the first definitions of $e$, beginning in 1618 when studying the limit of compound interest into continous interest.
Thus,
$$\lim_{u\to\infty}u\ln\left(1+\frac1u\right)=\ln(e)=1$$
Try to look at the Taylor expansion of $\log(1+2^{-x})$, which states $\log(1+2^{-x})\sim 2^{-x}$ for extremely small values of $2^{-x}$. So, when $x$ goes to infinity, this term decreases extremely fast. And it balances the increase of $2^x$.
$$\lim_{x \to \infty}2^x\log{(1+2^{-x})} = \lim_{x \to \infty}{2^x2^{-x}\log{(1+2^{-x})} \over 2^{-x}} = \lim_{x \to \infty}{\log{(1+2^{-x})} \over 2^{-x}} = \lim_{2^{-x} \to 0}{\log{(1+2^{-x})} \over 2^{-x}} = 1$$
I used $\lim_{x \to 0} {\ln(x + 1)\over x} = 1$.
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le\log(x)\le x-1}\tag 1$$
Using $(1)$ with $x$ replaced with $1+2^{-x}$ reveals that
$$\frac{2^{-x}}{1+2^{-x}}\le\log(1+2^{-x})\le 2^{-x} \tag2$$
Multiplying $(2)$ by $2^x$m we obtain
$$\frac{1}{1+2^{-x}}\le 2^x\log(1+2^{-x})\le 1 \tag3$$
whereupon applying the squeeze theorem to $(3)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}2^x\log(1+2^{-x})=1}$$
