How to prove or disprove that $\frac{\pi}{2}$ is limit of $\sin^{-1}x$ as $x\to 1$ by using $\epsilon-\delta$ method.

Question

How to prove or disprove that $\frac{\pi}{2}$ is limit of $\sin^{-1}x$ as $x\to 1$ by using $\epsilon-\delta$ method.

My attempt:

$|\sin^{-1}x-\frac{\pi}{2}|=|\cos^{-1}x|=|\sin^{-1}\left(\sqrt{1-x^2}\right)|=|\sin^{-1}\left(\sqrt{2(1-x)-(1-x)^2}\right)|<|\sin^{-1}\sqrt{2\delta-\delta^2}|$

If $\epsilon=\sin^{-1}\sqrt{2\delta-\delta^2}$

$\delta^2-2\delta+\sin^2\epsilon=0$

$\delta=1\pm\cos\epsilon$ or $\delta=1-\cos\epsilon$

So for every $\epsilon>0$ in the neighborhood of $0$ there does exist a $\delta>0$ and we are through.

If the above argument is correct then $$\lim_{x\to 1}\sin^{-1}x=\frac{\pi}{2}$$ But $\sin^{-1}x$ does not exist in the right neighborhood of $x=1$ i.e.$$\lim_{x\to 1^+}\sin^{-1}x=does\space\space not \space exist$$.

Why is there a contradiction in the two approaches

Answer 1

The contradiction you perceive is rooted in a looseness in the definition of limits.

A formal definition of a limit should state the domain and range of the function $f$ very explicitly, should restrict $a$ to be a limit point of the domain of $f$, and should restrict $x$ to vary only in the domain of $f$ (minus the point $a$, if it happens to be in the domain of $f$). When you do that, you get the following definition, with the bold faced portion emphasizing how $x$ is restricted:

Definition: Given a function $f : A \to B$ with $A,B \subset \mathbb R$, given $a \in \mathbb R$ which is a limit point of the set $A$, and given $L \in \mathbb R$, to say that $\displaystyle\lim_{x \to a} f(x) = L$ means that for every $\epsilon > 0$ there exists $\delta > 0$ such that for every $x \in A - \{a\}$, if $|x-a|<\delta$ then $|f(x)-L| < \epsilon$.

You can see a definition formulated in this manner in Fitzgerald's Advanced Calculus book.

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Regarding your proof, it seems a little convoluted and hard for me to follow, although I think the general idea is correct.

I would suggest a more direct approach, instead of trying to use an identity involving $\cos^{-1}$: Given $\epsilon>0$ let $\delta = 1-\sin(\pi/2-\epsilon)$ and then apply the fact that the sine function is continuous and strictly increasing on the domain $[-\pi/2,\pi/2]$.