On the domain of $f(x)=\sqrt{x+2} +\frac{1}{\log_{10}(1-x) }$

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.

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.