How would I calculate the limit of $g(x)$ that inside another function?

Question

So I was tutoring a friend today, and she had a interesting problem that she found in her textbook. The problem was like this (yes, I've changed the problem around for the sake of asking):

If $$\lim_{x \to 1} \frac{4-g(x)}{1-x} = 3$$ what is $\lim_{x \to 1} g(x)$?

My first thought was to jump straight in with the quotient rule but that requires the denominator to $\neq 0$ but then after that I was thinking about maybe theres a trick that you would have to multiply the top and bottom by to get this answer. Like for example: \begin{align} \lim_{x \to 1} \frac{4-g(x)}{1-x} &= 3 \\ \lim_{x \to 1} (4-g(x)) &= \lim_{x \to1} 3(x-1) \\ \lim_{x \to 1} (4-g(x)) &= 0 \end{align} giving that $\lim_{x \to 1} h(x) = 4$.

But I was unsure if this was the right way to do it. If anyone can offer a suggestion that would be very helpful

Answer 1

Since

$$\lim_{x\to1} \frac{4-g(x)}{1-x} = 3$$

and the denominator tends to $0$ it is necessarly

$$\lim_{x\to1} g(x)=4$$

otherwise the limit wouldn't exist finite.

Remarks after the comment by Yanko:

• given that $\lim_{x\to1} g(x)=4$ it is obviously not necessarily true that $\lim_{x\to1} \frac{4-g(x)}{1-x} = 3$ or that theat limit exist (e.g. $g(x)=4+|x-1|$).