Why is $\lim\frac{x^2\sqrt{1-x}-18}{(x^2-9)\log(x+4)}$ different for $x\to -3^{+}$ and $x\to -3^{-}$?

Question

$$\lim_{x\to-3}\frac{x^2\sqrt{1-x}-18}{(x^2-9)\log(x+4)}$$ I used Taylor expansion of $\sqrt{1-x}$ and $\log(x+4)$ centered in $x=-3$: $$\lim_{x\to-3}\frac{x^2(2)-18}{(x^2-9)(x+3)}=\lim_{x\to-3}2\frac{x^2-9}{(x^2-9)(x+3)}=\frac2{x+3}=+\infty$$

I thought that the two-sided limit $=+\infty$, but Wolfram Alpha gives: $$\lim_{x\to-3^-}f(x)=-\infty,\qquad \lim_{x\to-3^+}f(x)=+\infty$$

I thought of the following as an explanation for the different values: $$\text{if } x\to-3^-, x=-3-\varepsilon\implies \frac2{-3-\varepsilon+3}=-\frac2{\varepsilon}=-\infty\\\text{if } x\to-3^+, x=-3+\varepsilon\implies \frac2{-3+\varepsilon+3}=+\frac2{\varepsilon}=+\infty$$ where $\varepsilon$ is an infinitesimal. Is my explanation correct?

Answer 1

You ask two different questions: one in the title, one in the body. I will answer the second and comment on the first one.

Your explanation is correct, however, the derivation of the first limit is not completely correct. You cannot just replace $\sqrt{1-x}$ by $2$, it may be too short Taylor expansion. It gives the correct answer here at the end just because the order of the denominator is higher. In general (e.g. if you remove $x^2-9$ or $\log(4+x)$), it will produce a wrong result.

How to calculate the limit: replace as you did $x=-3+\epsilon$ and expand $\log(1+\epsilon)$ and $\sqrt{4-\epsilon}$ to the first order Maclaurin expansion with the remainder $o(\epsilon)$ and cancel one $\epsilon$. You will get asymptotics something like $\frac{\text{const}}{\epsilon}$, from where you conclude as you did that the side limits are $\pm\infty$.

Answer 2

Infinitesimals, as long as you work in a classical framework, aren't good objects to reason with. Limits should be thought of as processes, rather than values : how does a function behave when this variable is made as close as possible to this value? This is why writing things like $\frac{1}{+\infty}$ will be a disservice in the long run, as intuitive as it may look, as there are countless ways to approach infinity via limits.

Now that this is out of the way, yes, you're right, this is basically the same as with $ f(x) = \frac{1}{x} $ at $ 0 $

You have $ f(x) \underset{x \rightarrow 0^+}{\rightarrow} + \infty $ and $ f(x) \underset{x \rightarrow 0^-}{\rightarrow} - \infty $, because those are the limits of the function that you hopefully have computed in your classes. No need to introduce wonky objects, instead let $ \epsilon = x+3$ and just state that $ x \rightarrow -3^+ $ is equivalent to $ \epsilon \rightarrow 0^+ $. This is just nitpicking but in my opinion that's the difference between a vague explanation and a precise one.