Show that the expressions $\sin^{-1}(\frac{1}{\sqrt{x}})$ and $\frac{1}{\sqrt{x}}$ are the same for big values

Question

How can you show that the expressions $\sin^{-1}(\frac{1}{\sqrt{x}})$ and $\frac{1}{\sqrt{x}}$ for big values are the same?

The opposite side of a triangle is given with $1/\sqrt(x)$, the angle between the hypotenuse and the the opposite side can be calculated by $sin^{-1}(1/\sqrt(x))$. For large x that seems to be correct, as a review with some inserted values in the calculator has given. Can one also show this connection differently? Maybe graphically?

Answer 1

If $x$ is large, then $\frac1{\sqrt x}$ is small and if $y$ is small, then $\arcsin(y)$ is approximately $y$ (because $\arcsin(0)=0$ and $\arcsin'(0)=1$). But they are never equal if $y\neq0$.

Answer 2

Not that: $\lim_{x\rightarrow\infty}(\frac{1}{\sqrt{x}})=0$
$\Rightarrow \lim_{x\rightarrow\infty}(\sin^{-1}(\frac{1}{\sqrt{x}}))=\sin^{-1}(0)=0$

Hence: $\lim_{x\rightarrow\infty}(\frac{1}{\sqrt{x}})=\lim_{x\rightarrow\infty}(\sin^{-1}(\frac{1}{\sqrt{x}}))=0$

Answer 3

Consider the limit of their ratios: $$\lim_\limits{x\to +\infty} \frac{\arcsin \frac1{\sqrt{x}}}{\frac1{\sqrt{x}}}\stackrel{\frac1{\sqrt{x}}=t}{=}\lim_{t\to 0^+}\frac{\arcsin t}{t}\stackrel{L'H}=\lim_{t\to 0^+}\frac{\frac1{\sqrt{1-t^2}}}{1}=1.$$