Prove. $\lim_{x \to 2} \frac{x+3}{x-1} = 5$, from first princibles. I am a student and I'm practising proving limits from first principles. I've been asked the question above, I have attempted an answer, but any tips or tricks anyone has to prod me in the right direction would be much appreciated!
Firstly I fix $\epsilon > 0$ and fing $\delta > 0$ such that.
$$0< |x-2| < \delta \implies \bigg| \frac{x+3}{x-1} -5 \bigg|<\epsilon$$
$$\impliedby \bigg|-\frac{4(x-2)}{x-1} \bigg|$$
$$\impliedby \bigg|\frac{4(x-2)}{x-1} \bigg|$$
$$\impliedby \frac{4}{|x-1|}|x-2| $$
Let $|x-2| < 1$ then
$-1 < x-2 < 1$ or $0<x-1<2$
Let $\epsilon >0$ be given.
Choose $\delta = min(1,\frac{\epsilon}{2}$)
Then  $|x-2|< \delta \implies \frac{4}{|x-1|}|x-2| < 2\delta \leq \epsilon  $ 
Thanks for your time!
 A: The solution is almost correct, but the $0<x-1<2$ implies that $x-1$ can be really close to $0$, which is not good because it means that $\frac{1}{2} < \frac{1}{x-1} < +\infty$. So you need to squeeze it even more.
A: $x-1>0$ gives $\frac 4{x-1}<+\infty$ you need to be more restrictive.
My advice is to work with $x=2+u$ with $|u|<\delta$ rather than with $|x-2|$, it is more natural.
Thus $\left|\dfrac{x+3}{x-1}-5\right|=4\left|\dfrac{u}{1+u}\right|$ 
And now you see more clearly that you need something like $|u|<\frac 12$ to get $|1+u|>\frac 12$ and $\left|\frac 4{1+u}\right|<8$
It is exactly the same as your proof but using $u$ triggers more automatic reflexes than manipulating $|x-2|$. Ithink it is a good habit to take.
A: You can impose $\delta < \frac{1}{2}$, so you get $0<|x-2|<\delta<\frac{1}{2}$ and hence $x-1>\frac{1}{2}>0$; this means that $x-1>0$ and so $x-1=|x-1|$.
So you have $|x-1|=x-1>\frac{1}{2}$, hence $\frac{1}{|x-1|}<2$ and you can conclude taking $\delta=\min\{\frac{1}{2},\frac{\varepsilon}{8}\}$.
