Need help with finding the limit of a function I can't think of a way to solve this one. Any help or direction towards a solution is appreciated!

If $\lim_{x\to 1} \frac{f(x)-8}{x-1} = 10$, find $\lim_{x\to 1} f(x)$.

Thanks in advance!
P.S. Not any homework, just learning calculus on my own.
 A: Observe that
$$
f(x)-8=\frac{f(x)-8}{x-1}\cdot (x-1).
$$
And then find the limit of the left hand side as $x\to 1$.
A: Assume that it is not true that $\lim_{x\to 1}f(x)=8$. Then there exists a positive number $\epsilon_0>0$ such that for every $\delta>0$ we can find a number $x$ such that $0<|x-1|<\delta$ but $|f(x)-8|\geq\epsilon_0$. Therefore, for every $n$ we can find $x_n$ such that  $0<|x_n-1|<1/n$ but $|f(x_n)-8|\geq\epsilon_0$. In other words, there is a sequence $x_n$ converging to $1$ such that
$$\frac{|f(x_n)-8|}{|x_n-1|}\geq\frac{\epsilon_0}{1/n}=n\epsilon_0\to\infty$$
and this contradicts the assumption. Therefore, our assumption that the limit of $f(x)$ as $x\to 1$ is not $8$ leads to a contradiction, so it must be the case that $\lim_{x\to 1}f(x)=8$.
A: On direct substitution, we see that the denominator is $0$. For the limit to be $10$, the numerator should also be $0$.
Thus,
$$\lim_{x\rightarrow 1} \left[f(x)-8\right] =0$$
$$\implies \lim_{x\rightarrow 1} f(x) = 8$$
Hope this helps :)
Ask if you are confused at any step!
