mjhjmtu

Prove that $$lim_xto3(4x-5)=7$$

Then you get $$0<left|x-3right|<delta$$ $$left|x-3right|<fracepsilon4$$

Now we see $$delta=epsilon/4$$

Proof: Given $epsilon>0$, choose $delta=epsilon/4$. If $0<left|x-3right|<delta$, then $$left|(4x-5)-7right|=left|4x-12right|<4delta=4left(cfracepsilon4right)=epsilon$$

Thus$$\$$
if $0<left|x-3right|<delta$ then $left|(4x-5)-7right|<epsilon$
$$\$$QED

So my question is why under the word proof is $left|4x-12right|<4delta$?
Why is it $4delta$ and not just $delta$?

Because you started with $$0lt|x-3|ledelta$$
Multiply every term in this inequality by $4$ you get $$4cdot 0lt4cdot |x-3|le4cdotdelta$$
or, just using the last two terms $$|4x-12|lt 4delta$$

$$|4x-12|=4|x-3|$$

we have $|x-3| < delta$, hence

$$|4x-12|=4|x-3|< 4 delta$$

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 $$|x-3|<delta implies |f(x)-7|=|(4x-5)-7|=|4x-12|=4 |x-3|<4delta.$$ So, given $epsilon,$ if $delta =epsilon/4$ then $0<|x-3|<delta implies |f(x)-7|<4delta =epsilon.$

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 $0<|x-3|<delta'implies |f(x)-7|<epsilon.$