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Let, $$ f(x) =\sqrt{x+2}+\frac{1}{\log_{10}(1-x) }$$ Then my textbook mentions that domain of $f$ is $[-2, 1)\setminus \{0\}$. It proves that fact by considering $$(x+2) \gt 0$$ $$(1-x) \gt 0$$ Now I argue that since we are speaking under the context of real numbers and real analysis it's allowed to take $(1-x) \ge 0 $. Because $$\lim_{x\to 1}f(x) =\sqrt{3}\in\mathbb{R}$$ So I claim that domain of $f$ should be $[-2, 1]\setminus \{0\}$. I request a clarification if my claim is wrong.

Vishal Uchiha
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  • The limit at $x=1$ exists, even though the function is not defined there. So it should be an open interval at $1$ – Andrei Dec 16 '21 at 08:25
  • Similar questions: https://math.stackexchange.com/q/713069/42969, https://math.stackexchange.com/q/3320936/42969 – Martin R Dec 16 '21 at 08:27
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    I would say that the function is not defined there, even though the limit at $x = 1$ exists. :-P – Brian Tung Dec 16 '21 at 08:31

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I agree with @BrianTung. The given definition of $f$ doesn't work at $x=1$, even though the one-sided limit $\lim_{x\to1^\color{blue}{-}}f(x)$ exists. What you've discovered is a removable left-discontinuity; we can modify the definition so $f(1)=\sqrt{3}$, thus making $f$ left-continuous at $1$. However, this is no longer your original function, just an extension of it, so ideally it needs a different symbol.

J.G.
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  • Actually, in the context of this question, the one-sided limit is equal to the actual limit, simply because there is no meaningful notion of a right-sided limit to investigate. Limits are only meaningful to consider for the limit points of the domain, in the topological sense, and it only makes sense to investigate them by quantifying over points in the domain. – Angel Dec 16 '21 at 13:52
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  1. The radical expression must be greater than or equal to zero $$x+2\geq 0\Rightarrow x\geq -2$$
  2. We have a fraction therefore the denominator is not zero $$\log_{10}\left ( 1-x \right )\neq 0\Rightarrow 1-x\neq 1\Rightarrow x\neq 0$$
  3. Since the denominator is the logarithm, therefore, its argument will be strictly greater than zero $$1-x>0\Rightarrow x<1$$ Then the scope of the function will be: $$x\in \mathbb{R} \; : \; x\in \left [ -2,1 \right )\setminus \left \{ 0 \right \}$$
Fk-91
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The fact that $$\lim_{x\to1}f(x)=\sqrt{3}$$ does not imply that $$f(1)=\sqrt{3},$$ this is not how limits work. For example, I can have a piecewise definition of $f$ such that $\begin{cases}\frac{x}{x}&x\neq0\\0&x=0\end{cases}.$ With this definition, we have that $$\lim_{x\to0}f(x)=1,$$ yet $$f(0)=0.$$ In general, unless the function is continuous at $p,$ there is no relationship at all between $$\lim_{x\to{p}}f(x)$$ and $f(p).$ And the fact that you can evaluate the limit for a point does not imply that point is in the domain. In fact, it can be possible for a point to be in domain, and yet for the limit to not even exist. So, with all due respect, your claim is inaccurate.

Angel
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