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Let us consider the following limit:

$$\lim_{x\to\infty} \operatorname{cosec}^{-1}\frac{x}{x+7}.$$

According to my understanding both $x$ as well as $x+7$ tend toward $\infty$ as they can be cancelled to give us $1$ and the limit becomes $\pi$/$2$.

But in reality, the limit doesn't exist could someone explain why?

kipf
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TheCuriousOne
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  • Actually it is not correct, @kipf - take a look at the graph and consider the function's domain. – PrincessEev Jul 10 '23 at 07:46
  • begin by asking yourself the question: for which values of $u$ does $\csc^{-1} u$ exist? – David Raveh Jul 10 '23 at 07:49
  • $\csc(\theta) = \dfrac{1}{\sin(\theta)},~$ and $~\sin(\theta)~$ can not be larger than $~+1.$ – user2661923 Jul 10 '23 at 07:51
  • You might want to check values of $\operatorname{cosec}^{-1}\frac{x}{x+7}$ for some arguments first, say, $x = 1, \ 20, \ 7000, \ 10^{10},$ and see what they suggest... :) – CiaPan Jul 10 '23 at 09:09

1 Answers1

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Your function will need to be defined.

Note that the function $ \operatorname{arccsc}(x) $ has domain $(-\infty,-1] \cup [1,\infty)$.

However, $$ x > 0 \implies x+7 > x \implies 0 < \frac{x}{x+7} < 1 $$ so $\operatorname{arccsc}(\frac{x}{x+7})$ is not defined as $x \to \infty$. It certainly doesn't have a limit as a result.

Your error is trying to use continuity and say $$ \lim_{x \to \infty} \operatorname{arccsc}\left (\frac{x}{x+7} \right) = \operatorname{arccsc}\left (\lim_{x \to \infty} \frac{x}{x+7} \right) $$ This makes sense for functions continuous at and around a finite value $c$ as $x \to c$, but one has to be careful as $x \to \infty$. (For instance, "continuous at infinity" generally is a nonsensical notion.) As one sees here, that case can be quite a bit more complicated.

PrincessEev
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