I am trying to explain why the following function is not well defined and also have a question had it been a one-sided limit approaching infinity on a restricted domain.
So,
$$
\lim _{x \rightarrow -\infty} \sin ^{-1}\left(\frac{2-x}{x}\right)
$$
$$
\lim _{x \rightarrow-\infty} \sin ^{-1}\left(\frac{\frac{2}{x}-\frac{x}{x}}{\frac{x}{x}}\right) = \sin ^{-1}(-1) = -\pi/2
$$
$$\lim _{x \rightarrow-\infty^-} = undefined$$ $$\lim _{x \rightarrow-\infty^+} = -\pi/2$$
$x<-1$ is outside of the domain. Therefore, the limit is not well-defined because as $\left(\frac{2-x}{x}\right)$ approaches $-1$ from the left, $\sin ^{-1}(x)$ is not defined.
Have I got that right?
And my other question is had the original question been:
$$
\lim _{x \rightarrow -\infty^+} \sin ^{-1}\left(\frac{2-x}{x}\right)
$$
Then would the answer have been – it exists but as it is a one-sided limit it can't be well defined?
Thanks.
Best Answer
When dealing with limits approaching infinity, they are by definition only one-sided. That is,
\begin{align*} \lim_{x\to\infty}f(x)&=\lim_{x\to\infty^-}f(x)\\ \lim_{x\to-\infty}f(x)&=\lim_{x\to-\infty^+}f(x) \end{align*}
So, $\lim_{x\to-\infty^-}$ has no meaning. The issue here is that we are trying to have $x\to-\infty$, but $\sin^{-1}\left(\frac{2-x}{x}\right)$ is not defined for $x\in(-\infty,1)$. To see this, recall that the domain of $\sin^{-1}(x)$ is $-1\leq x\leq1$. So, for $\sin^{-1}\left(\frac{2-x}{x}\right)$, we need $-1\leq\frac{2-x}{x}\leq1$.
If $x<0$, then
\begin{align*} -1\leq\frac{2-x}{x}\leq1&\implies -x\geq2-x\geq x\\ &\implies 0\geq2\geq 2x\\ &\implies 0\geq 1\geq x. \end{align*}
So if $x<0$, then we can conclude that $0\geq 1$, which is impossible. So, $x\not<0$. This is really enough to show that $x\to-\infty$ is not possible, but lets keep going.
If $x>0$, then
\begin{align*} -1\leq\frac{2-x}{x}\leq1&\implies -x\leq2-x\leq x\\ &\implies 0\leq2\leq 2x\\ &\implies 1\leq x. \end{align*}
So we have $x>0$ and $x\geq 1$, which is the same as just $x\geq 1$. Hence, our domain is $[1,\infty)$, and taking the limit as $x\to-\infty$ is not possible.