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Find the following limit: $\lim \limits_{x \to -1^{-}} \arccos(x)$, where $\arccos(x)$ means $\cos^{-1}(x)$.

My main question is what imaginary value does this limiting value takes?

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    Can someone answer this question? – 2017-02-13
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    Perhaps someone can. Anyway, the **real** function $\;\arccos x\;$ is not even defined to the left of $\;-1\;$ and thus, to make some sense of this, one could assume this is *the complex* $\;\arccos\;$, which is a real multivalued mess with the complex logarithm and/or square roots of stuff. Perhaps the limit doesn't even exist... – 2017-02-13
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    Okay x< -1 cos inverse doesn't exist but still the value of limit could be calculated irrespective of the fact that the value is imaginary complex number – 2017-02-13
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    @Ma I think you're missing my point: **if** you want to do *real calculus* then you want to evaluate the limit of a non-existing thing, and this cannot be done. Period. If you want to get into complex analysis then, imo, you first would need to know the explicit form of the complex inverse cosine, which is a very non-pretty thing. Here, take a peek: http://scipp.ucsc.edu/~haber/archives/physics116A10/arc_10.pdf , pages 6-7 . – 2017-02-13
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    You need to specify how the arccosine function is being extended beyond its "natural" domain, $[-1,1]$. – 2017-02-13
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    Antonio sir can you guide me as to which book should I refer in calculus. – 2017-02-13
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    @MayankJha To pin someone you must add @ before his/her nick. Anyway, there're lots of excellent calculus books: Spivak, Lang, Stewart, Thomas, Rudin etc. Google "calculus" and check some titles. What you want here though is more advanced: complex calculus, which you can also google: Alhfors, Remmert, Ullrich, Rudin, Conway, etc. – 2017-02-13

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Well, we can't really approach the function $f: \mathbb{R} \to \mathbb{R}, x \to \arccos(x)$ from the negative direction, because it is not defined. We can change this to be $f: \mathbb{C} \to \mathbb{C}, x \to \arccos(x)$ and we can approach from a direction where $\Im(f)=0$ though. If we express $f$ in complex form we find that $$\arccos(z) = -i\log(z+\sqrt{z^2-1})$$ If we stick to the appropriate branch of the complex logarithm we can say that, for all $x<-1$, $\Re(f) = \pi$. Moreover, we can define $\Im(f)=-\log \left|z+\sqrt{z^2-1}\right|$ which has limit $\lim_{z \to 0} \Im(f) = 0$

If an image would help, here you go. da image