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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?

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    $\begingroup$ The wording is important - for large values they are approximately the same. Note that the Taylor expansion for arcsin involves a sum of odd powers, and for small values, those powers become negligible. $\endgroup$ – TheSimpliFire Mar 16 '19 at 8:00
  • $\begingroup$ Hint: $\sin'(0)=1$. $\endgroup$ – Wolfgang Kais Mar 16 '19 at 8:04
  • $\begingroup$ @TheSimpliFire if I understand you right, you mean: $sin^{-1}(x)=x-\frac{1}{6}x^3+\frac{3}{40}x^5+...$. Then this means for $sin^{-1}(1/\sqrt{x})$ something like that: $1/\sqrt(x)+-\frac{1}{6}(1/\sqrt(x))^3+...$ right? $\endgroup$ – P_Gate Mar 16 '19 at 8:16
  • $\begingroup$ Yes, and when $x\to\infty$, $1/\sqrt x\to0$ so powers of three, five etc become extremely small. $\endgroup$ – TheSimpliFire Mar 16 '19 at 8:20
  • $\begingroup$ If you agree with the approximation $\sin\theta\approx \theta$ when $\theta\to 0$, then the result you seek is just considering $\theta:=1/\sqrt x\to 0$ as $x\to\infty$ and applying $\arcsin$ on both sides. $\endgroup$ – learner Mar 16 '19 at 8:24
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Consider the limit of their ratios: $$\lim_\limits{x\to +\infty} \frac{\arcsin \frac1{\sqrt{x}}}{\frac1{\sqrt{x}}}\stackrel{\frac1{\sqrt{x}}=t}{=}\lim_{t\to 0^+}\frac{\arcsin t}{t}\stackrel{L'H}=\lim_{t\to 0^+}\frac{\frac1{\sqrt{1-t^2}}}{1}=1.$$

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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 $y\neq0$.

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

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