Prove that $ \lim_{x\to\infty} 2^x\cdot \arcsin{\dfrac{1}{2^x}} = 1$ without L'Hopital 
Prove that $ \lim_{x\to\infty} 2^x\cdot \arcsin{\dfrac{1}{2^x}} = 1$.

This is easy to do with L'Hopital (by converting it to $0/0$ form).
Is there any way to do it without using it?
I have no clue where to start.
 A: Another substitution besides the one in the comments: $u = \arcsin(1/2^x)$ turns the limit into $$\lim_{u \to 0} \frac{u}{\sin u},$$ and this limit or its reciprocal is usually calculated early on in calculus to be equal to $1$, as we can't use the limit definition of the derivative to differentiate sine or cosine without knowing that $\lim_{h \to 0} \frac{\sin(h)}{h} = 1$.
A: $$L\lim_{x\to\infty} 2^x\cdot \arcsin{\dfrac{1}{2^x}} = \lim_{u\to 0} \dfrac  {\arcsin u}u $$
$$L=\lim_{t\to 0} \dfrac t {\sin t } $$
Now note that:
$$\lim_{t\to 0} \dfrac {\sin t }t=\lim_{t\to 0} \dfrac  {\sin t -\sin 0 }{t-0} =(\sin t)'|_{t=0}=\cos 0=1$$
$$\implies L=\dfrac 1 {\lim\limits_{t\to 0} \dfrac {\sin t } t}=1$$
Where $u=\dfrac 1 {2^x}$ and $u=\sin t$.
A: Step 1: Substitute $y = \dfrac 1 {2^x}$, you get $\lim_{y \to 0} \dfrac{\arcsin y}{y} = 1$
Step 2: Substitute $z = \arcsin y$, you will get $\lim_{z \to 0} \dfrac{z}{\sin z} = 1$. Does it look familiar now?
A: Let $t=\frac{1}{2^x}$ and $g(t)=\arcsin(t)$, then we have
$$ \lim_{x\to\infty} 2^x\cdot \arcsin{\dfrac{1}{2^x}} = \lim_{x\to\infty} \frac{ \arcsin{\dfrac{1}{2^x}}}{\frac{1}{2^x}} $$
$$=\lim_{t\rightarrow 0}\frac{\arcsin(t)}{t}=\lim_{t\rightarrow 0}\frac{\arcsin(t)-\arcsin(0)}{t-0}=g'(0)=1.$$
Or you can use the series expansion $\frac{\arcsin(t)}{t}=1+\frac{t^2}{6}+O(t^4).$
