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The problem goes: is it possible to extend the function to a continuous on the entire $\mathbb{R}$(: $$f(x)=\arctan{\frac{1}{x-1}}\\$$

Ok my logic is the following: looking at the graph of $arctan(x)$ I think it is continuous on the entire $\mathbb{R}$, so I assume the only problem is the expression $\frac{1}{x-1}$, and therefore the function is discontinuous only for $x=1$.$\\$

Now I assume that I should assign the value for $x=1$ to $f(x)=0$, since the expression $\frac{1}{x-1}$ can't amount to 0, so that value will be skipped on the graph.Therefore, my idea was to do this: $$f(x)\begin{cases} 0 ;x=1\\ \arctan{\frac{1}{x-1}} ; x\neq1 \\ \end{cases}$$

Since the value of $arctan(0)$ equals to $0$. Now, defined this way the function should be the same as $arctan(x)$ and is therefore continuous on the entire $\mathbb{R}\\$. The problem is in the section "limits without L'hospital" so I'm not really allowed to use derivations or integrals to help with this problem.

Please do correct me if I made mistakes! Thanks in advance!!

2 Answers 2

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The map $f$ is well defined and continuous on the domain $\mathbb{R}-\{1\}$.

It is known that :

$$\lim_{t\to+\infty}\arctan(t)=\frac{\pi}{2}\quad\mathrm{and}\quad\lim_{t\to-\infty}\arctan(t)=-\frac{\pi}{2}$$

It follows that :

$$\lim_{x\to1^-}f(x)=-\frac{\pi}{2}\quad\mathrm{and}\quad\lim_{x\to1^+}f(x)=\frac{\pi}{2}$$

So $f$ cannot be "extended" (that's probably the word you were looking for) into any continuous map defined on $\mathbb{R}$.

Recall that given $\alpha\in\mathbb{R}$ and a map $f:\mathbb{R}-\{\alpha\}\to\mathbb{R}$, the continuity of $f$ at $\alpha$ is equivalent to the existence of finite left and right limits at $\alpha$, those limits beeing moreover equal.

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    Yes, I guess I went in the totally wrong direction! Thanks a lot!! – 2017-01-29
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The key is $\lim_{x\to\pm\infty}\arctan{x}=\pm\pi/2$

What is the limit of $f$ when $x\to 1^{+}$: prove it is $\pi/2$

The limit of $f$ when $x\to 1^{-}$ is $-\pi/2$ (Prove it) and so there is no continuation at $x=1$ such that $f(1)=\lim_{x\to 1^{+}} f(x)=\lim_{x\to 1^{-}} f(x)=f(1)$