Compute $\lim_{x \to -1} f(x)$ for a function $f: \mathbb R \to \mathbb R$ such that
$$4 = \lim_{x \to -1} \frac{f(x)+2}{x+1} - \frac{x}{x^2-1} \tag{1}$$
$$ = \lim_{x \to -1} \frac{f(x)+2}{x+1} - \frac{\frac{x}{x-1}}{x+1}$$
$$ = \lim_{x \to -1} \frac{f(x)+2 - \frac{x}{x-1}}{x+1}$$
- Solution 1: My approach is that the numerator $f(x)+2 - \frac{x}{x-1}$ must approach zero as $x \to -1$ because the denominator approaches zero as $x \to -1$ and so $\lim_{x \to -1} f(x) = -\frac{3}{2}$.
Are we allowed to do the following, which seems to be the required solution, instead? This does not seem very rigorous, and I have a feeling there are obvious counterexamples. Of course, we can use $\varepsilon-\delta$ to check our answer, but I would like to know if and how this can be generalised for any function $f: \mathbb R \to \mathbb R$.
- Solution 2: Observe
$$0 = \lim_{x \to -1} x+1$$
Then
$$4 = \lim_{x \to -1} \frac{f(x)+2 - \frac{x}{x-1}}{x+1}$$
$$\implies 0 = 0 \cdot 4 = (\lim_{x \to -1} x+1) \cdot 4 = \lim_{x \to -1} \frac{f(x)+2 - \frac{x}{x-1}}{x+1} \lim_{x \to -1} x+1$$
$$ = \lim_{x \to -1} \frac{f(x)+2 - \frac{x}{x-1}}{x+1} x+1 = \lim_{x \to -1} (f(x)+2 - \frac{x}{x-1})$$
$$ = \lim_{x \to -1} f(x) + \lim_{x \to -1} 2 - \lim_{x \to -1} \frac{x}{x-1} = \lim_{x \to -1} f(x)+2 - \frac12 = \lim_{x \to -1} f(x)+ \frac32$$
$$\implies - \frac32 = \lim_{x \to -1} f(x) \tag{2}$$
I suspect the preceding solution is not rigorous, and this is a case where we have only a guess and must check our answer by proving $(1)$, by computation since $\varepsilon-\delta$ is actually not yet allowed, assuming $(2)$, so if one does then preceding solution, then one must follow up with a computation. I'm not quite sure what the problem is, but it might be that we don't know $\lim_{x \to -1} f(x)$ exists in the first place.
Is the fact that the domain and range are both $\mathbb R$ relevant? I think of a counterexample like $f: \{7,8,10\} \to \{1,2\}$ or $f: C \to \mathbb Q^c$ where $C$ is the Cantor set.
If $l:=\lim_{x\to a}g(x),\,m:=\lim_{x\to a}h(x)$ exist and aren't $\pm\infty$, you can indeed write $lm:=\lim_{x\to a}g(x)h(x)$. In the case $l=0$ note that, when $x$ is close enough to $a$ to constrain $h$ to some interval around $m$, this upper-bounds $|h|$ (say with $|h|\ge U,\,U<\infty$) so $|g|\le\frac{\epsilon}{U}\to|g|\le\epsilon$. Note that as long as the domain and range are such that $|g|,\,|h|$ can be bounded and one is arbitrarily small in the limit, we're fine.