I was given this question:
If $$\lim_{x\to 1} \frac{f(x)-6}{x-1} = 10$$ then what is $$\lim_{x\to 1} f(x)$$
I am assuming that I will need to use limit laws in reverse, but this doesn't seem to work as the limit of $x-1$ as $x$ approaches $1$ is $0$, which gives an undefined answer.
Let $\alpha: \Bbb{R} \setminus\{1\} \to \Bbb{R}$ be the function defined by \begin{align} \alpha(x) = \dfrac{f(x)-6}{x-1} \end{align} Then, by assumption, $\lim\limits_{x \to 1}\alpha(x) = 10$. Also, it's easy to see that $\lim \limits_{x \to 1}(x-1) = 0$. Hence, by the rules for products of limits, we know that $\lim \limits_{x \to 1} \big(\alpha(x) \cdot (x-1) \big)$ exists and: \begin{align} \lim \limits_{x \to 1} \bigg(\alpha(x) \cdot (x-1) \bigg) &= \bigg(\lim_{x \to 1} \alpha(x) \bigg) \cdot \bigg( \lim_{x \to 1}(x-1)\bigg) \\ &= (10) \cdot 0 \\ &= 0 \tag{$*$} \end{align}
But, now, let's examine the LHS more carefully. If $x \neq 1$, what is $\alpha(x) \cdot (x-1)$? Well, let's go back to the definition of $\alpha$. It should be clear that if $x \neq 1$, then $\alpha(x) \cdot(x-1) = f(x)-6$. Hence, what $(*)$ shows is that \begin{align} \lim_{x\to 1} \bigg( f(x) - 6 \bigg) = 0 \end{align} or equivalently, \begin{align} \lim_{x \to 1} f(x) = 6 \end{align}
The thought process I had when approaching this question is that we are interested in the limit of $f(x)$, and $f(x)$ appears in the numerator. Typically, I would use the rule \begin{align} \lim_{x \to 1} \bigg(\dfrac{\phi(x)}{\psi(x)} \bigg) = \dfrac{\lim\limits_{x \to 1} \phi(x)}{\lim\limits_{x \to 1} \psi(x)} \end{align} but in this case, we cannot directly apply this rule, because this rule is only valid when $\lim\limits_{x \to 1} \psi(x) \neq 0$ (division by zero is your worst enemy in math!). In this case, the denominator $x-1$ clearly has a limit of $0$ as $x \to 1$.
Therefore, to get around this obstacle, the most natural thing to do is simply "get rid of" the denominator by multiplying the denominator throughout, and as you can see, this was pretty much the idea behind my proof above.
Hopefully this helps you with not only the proof, but more importantly how to think about approaching such problems.
Since the limit as $x\to 1$ of the denominator is 0, the entire expression must explode to $\pm \infty$ if the limit of the numerator is nonzero. Therefore, since the limit exists, it must hold that $\lim_{x\to 1} (f(x) - 6)=0$. Thus $\lim_{x\to 1} f(x)=6$.
The limit will exist if $\lim_{x \to 1} f(x) = 6$. For example take $f(x) = 10x -4$ in which case, the given condition holds and the limit of the ratio is indeed $10$.
$$ \begin{align} \overbrace{\lim_{x\to1}\frac{f(x)-6}{x-1}}^{10}\overbrace{\vphantom{\frac61}\lim_{x\to1}(x-1)}^0 &=\overbrace{\lim_{x\to1}\frac{f(x)-6}{x-1}(x-1)}^{10\,\cdot\,0}\\ &=\lim_{x\to1}(f(x)-6) \end{align} $$
For any $\epsilon>0$, by definition, we can find $\delta_1>0$ such that $$\left|\frac{f(x)-6}{x-1}-10\right|<\epsilon \ \text{for all } 0<|x-1|<\delta_1,$$ So $$|(f(x)-6)-10(x-1)|<\epsilon|x-1| \ \text{for all } 0<|x-1|<\delta_1.$$ Let $\delta=\min\{\delta_1,\frac12,\frac{\epsilon}{20}\},$ then for all $0<|x-1|<\delta$, we have $$|f(x)-6|\leq |(f(x)-6)-10(x-1)|+10|x-1|<\epsilon\delta+10\delta\leq \frac12\epsilon+\frac12\epsilon=\epsilon.$$ By definition, we have $\lim_{x\to 1}f(x)=6$.
Just another way.
Making the problem more general and admiting that the function is well-conditioned, using Taylor series around $x=a$ $$f(x)=f(a)+ f'(a)(x-a)+\frac{1}{2} f''(a)(x-a)^2+O\left((x-a)^3\right)$$
$$\frac{f(x)-b}{x-a}=\frac{f(a)-b}{x-a}+f'(a)+\frac{1}{2} f''(a)(x-a)+O\left((x-a)^2\right)$$ So, to have $$\lim_{x\to a} \frac{f(x)-b}{x-a} = c$$ you need $f(a)=b$ and $f'(a)=c$.
It's possible if $$\lim_{x\rightarrow1}(f(x)-6)=0.$$ Can you end it now?
A full solution:
We have: $$\lim_{x\rightarrow1}f(x)=6+\lim\limits_{x\rightarrow1}(f(x)-6)=6+\lim\limits_{x\rightarrow1}\left(\frac{f(x)-6}{x-1}\cdot(x-1)\right)=$$ $$=6+\lim\limits_{x\rightarrow1}\frac{f(x)-6}{x-1}\cdot\lim\limits_{x\rightarrow1}(x-1)=6+10\cdot0=6.$$
