Definition of a limit, proving the limit is a certain value Prove that $$\lim_{x\to3}(4x-5)=7$$
Then you get $$0<\left|x-3\right|<\delta$$ $$\left|x-3\right|<\frac{\epsilon}{4}$$
Now we see $$\delta=\epsilon/4$$
Proof: Given $\epsilon>0$, choose $\delta=\epsilon/4$. If $0<\left|x-3\right|<\delta$, then $$\left|(4x-5)-7\right|=\left|4x-12\right|<4\delta=4\left(\cfrac{\epsilon}{4}\right)=\epsilon$$
Thus$$\\$$
if $0<\left|x-3\right|<\delta$ then $\left|(4x-5)-7\right|<\epsilon$
$$\\$$QED
So my question is why under the word proof is $\left|4x-12\right|<4\delta$?
Why is it $4\delta$ and not just $\delta$?
 A: $$|4x-12|=4|x-3|$$
we have $|x-3| < \delta$, hence 
$$|4x-12|=4|x-3|< 4 \delta$$
A: Because you started with $$0\lt|x-3|\le\delta$$
Multiply every term in this inequality by $4$ you get  $$4\cdot 0\lt4\cdot |x-3|\le4\cdot\delta$$
or, just using the last two terms $$|4x-12|\lt 4\delta$$
A: Let $f(x)=4x-5.$ The object is to prove that for any $\epsilon >0$ we can find some $\delta >0$  such that $|f(x)-7|<\epsilon$ whenever $0<|x-3|<\delta.$
In general, for a given value of $\epsilon,$ a value of $\delta$ that  works will depend on $\epsilon$ and on the nature of the function $f.$ And we do  not need to find the largest possible value of $\delta$ that will work.
The proof shows by elementary algebra that if $\epsilon >0$ and if $\delta=\epsilon /4$ then $ 0<|x-3|<\delta \implies |f(x)-7|<\epsilon.$ This can be discovered, rather than confirmed, by looking at the consequences of $|x-3|<\delta$ for $any$
$\delta$. We have $$|x-3|<\delta \implies |f(x)-7|=|(4x-5)-7|=|4x-12|=4 |x-3|<4\delta.$$ So, given $\epsilon,$  if $\delta =\epsilon/4$ then $0<|x-3|<\delta \implies |f(x)-7|<4\delta =\epsilon.$
So letting $\delta= \epsilon/4$ is sufficient. And it happens to be the largest value of $\delta$ that will work. But  we can also say that  if $\delta'=\epsilon /10^{10}$ then $0<|x-3|<\delta'\implies |f(x)-7|<\epsilon.$
