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$ \lim\limits_{x\rightarrow\frac{\pi}{2}^-} \tan x=+ ∞$ and $ \lim\limits_{x\rightarrow\frac{\pi}{2}^+} \tan x=- ∞$ , so $\tan\frac{\pi}{2}$ is not defined
$ \lim\limits_{x\rightarrow\frac{\pi}{2}^-} |\tan x|= \lim\limits_{x\rightarrow\frac{\pi}{2}^+} |\tan x|=+ ∞$.
So can we say that $\left|\tan\frac{\pi}{2}\right|=+ ∞$ and thus $|\tan x|$ is continuous at all points?

Asher2211
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  • $+\infty$ is not a known number. We only know that is a positive but can’t tell which one exactly. Continuity relates to a number we can determine. – WindSoul May 23 '21 at 13:53
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    No, since $\infty$ is NOT a real number, existence of limit being hampered – pmun May 23 '21 at 13:53
  • @WindSoul In my book the condition for continuity of f(x) at $x=a$ is left hand limit( at x=a)$=$right hand limit( at x=a)$=$$f(a)$. Here they don't mention $f(a)$ to be a finite value but for derivatives they mention that $f(a)$ is finite. So according to this definition shouldn't the limit exist? – Asher2211 May 23 '21 at 13:59
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    We have a continuous function $|\tan|\colon \Bbb R\to \Bbb R\sqcup {\infty}\cong \Bbb S^1$, but there is no continuous function $f\colon \Bbb R\to \Bbb R$ such that $f(x)=|\tan x|$ for all $x\not\in \left{\frac{n\pi}{2}:n\in \Bbb Z\right}$. The sign $\cong$ is for homeomorphism i.e., $\Bbb S^1$ is the one-point compactification of $\Bbb R$. – Sumanta May 23 '21 at 14:10
  • @Asher2211, when you say x=a you mean a known number. Infinity is not a known number. – WindSoul May 23 '21 at 21:29

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Just two comments:

  1. By writing "$ f(x) \rightarrow + \infty$ as $x \rightarrow a$" we express a particular way in which the limit as $x$ tends to $a$ does not exist: it expresses the fact that we can make the values of $f$ arbitrarily large making $x$ close enough to $a$.

  2. You can't talk about continuity of a function at a point in which the limit does not exist.

Amelian
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  • Isn't limit$\rightarrow∞$ and limit does not exist two different things? – Asher2211 May 23 '21 at 14:10
  • Check the definition of limit of a function; it talks about the values of a function being arbitrarily close to a (unique) real number by means of taking $x$ close enough to $a$; "$f(x) \rightarrow + \infty $ as $x \rightarrow a$" means that the values of $f$ increase without bound by means of taking $x$ close enough to $a$. Then to "tend to infinity" is a particular way of not converging. Don't let the word "tend" on the last expression mislead you ! @AsherAbraham – Amelian May 23 '21 at 14:14
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No, definitely not. A limit with an infinite limiting value does not exist. For continuity it is essential that both side limits exist, are equal, and are equal to the functional value at that point. It does not make much sense to equate infinities, as you have done.

Ritam_Dasgupta
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