Calculus – Limit of Arcsin and Arccos Functions as x Approaches 0

calculuslimits

I'm trying to evaluate this:
$$\lim_{x\to 0}\arcsin\left(\frac{\arccos\left(x\right)}{\pi}+\frac{\arccos\left(x^{2}\right)}{\pi}\right)$$

Now, calculating the for $$\lim_{x\to 0}\frac{\arccos\left(x\right)}{\pi}$$
This should equal to $\frac{1}{2}$
And similarly, $$\lim_{x\to 0}\frac{\arccos\left(x^2\right)}{\pi}$$
should equal to $\frac{1}{2}$.

And if I understand correctly, $\lim_{x\to 0}$ returns the actual value, not approximately, but the actual value that the function is approaching.

So, here $\frac{1}{2} + \frac{1}{2} = 1$

And, $$\arcsin\left(1\right) = \frac{\pi}{2}$$
which is my answer. Is this valid?
The answer given in my textbook is that the limit is not defined. But, I don't understand that.

Best Answer

The left hand limit does not exists in the ordered field $\mathbb{R}$ with the standard domain of the function $\left[-1,1\right]$. However, the limit exists in the complex plane $\mathbb{C}$ and in WolframAlpha $\arcsin(z)$ is defined for in the whole complex plane.

What does this means?

In the section "Background & Context here we read :

  • ArcSin in WolframAlpha is defined for complex argument via

$$ \sin^{-1}(z)=-i\log\left(\sqrt{|1-z^2|} e^{-\frac{1}{2}i\arg(1-z^2)}+iz\right)$$

  • ArcSin[z] has branch cut discontinuities in the complex plane.

However, despite the discontinuities in the section "Scope->Function Properties" we read:

  • ArcSin is defined for all real values from the interval $\left[-1,1\right]$.
  • Complex domain is the whole plane.

This means that we can calculate limits outside the real domain $\left[-1,1\right]$

Suppose that we want to calculate the limit of $\arcsin(z)$ approaching $z=1$ through real numbers at the right. See what is happening:

$$\arcsin(1.1) \approx 1.5708 - \color{red}{0.443568 i}$$ $$\arcsin(1.01) \approx 1.5708 - \color{red}{0.141304 i}$$ $$\arcsin(1.001) \approx 1.5708 - \color{red}{0.0447176 i}$$ $$\arcsin(1.0001) \approx 1.5708 - \color{red}{0.014142 i}$$ $$\arcsin(1) \approx 1.5708 \approx \frac{\pi}{2}$$

The values of $\arcsin(z)$ are approaching $\displaystyle \frac{\pi}{2}$ through the complex numbers not the reals.

An example with your function if we suppose we are approaching $z=0$ through the real numbers at the left:

$$ \arcsin\left(\frac{\arccos\left(-0.1\right)}{\pi}+\frac{\arccos\left((-0.1)^{2}\right)}{\pi}\right) \approx 1.5708 - \color{red}{0.0793683 i}$$

$$ \arcsin\left(\frac{\arccos\left(-0.001\right)}{\pi}+\frac{\arccos\left((-0.001)^{2}\right)}{\pi}\right) \approx 1.5708 - \color{red}{0.025218 i}$$

$$ \arcsin\left(\frac{\arccos\left(-0.0001\right)}{\pi}+\frac{\arccos\left((-0.0001)^{2}\right)}{\pi}\right) \approx 1.5708 - \color{red}{0.00797843 i}$$

So, again: the limit in the reals does not exists with the real domain $\left[-1,1\right]$, only in the complex plane with the domain extended to the whole plane, which is the definition used by Wolfram.

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