Epsilon delta proof for $\lim_{x \to 1} (5x-3) = 2$

Question

I cannot understand if proof is worded correctly:

Let $x \in (2-\delta, 2+\delta)$. Then we have that $f(x)=5x-3=2+3(x-1)$. Then $f(x) \in (2-3\delta, 2+3\delta)$

Chosing $3\delta = \epsilon$ we see that $f(x) \in (2-\epsilon, 2+\epsilon)$. Thus for $\delta $ such that $0 < |x-1| < \delta$ we have $\epsilon$ such that $0<|f(x)-2|<\epsilon$. So the limit is $2$.

What is wrong with the wording of proof above? I does not understand how to go from $x \in (2-\delta, 2+\delta)$ to $f(x) \in (2-3\delta, 2+3\delta)$. What have I done, is it correct?

Also my book Calculus - Thomas and Finney says that we can choose any $\delta \le \frac{\epsilon}{5}$, but in my case it is $\frac{\epsilon}{3}$

Answer 1

Let $\epsilon>0$.

We need $$|5x-3-2|<\epsilon$$ or $$|x-1|<\frac{\epsilon}{5}.$$

Thus, for all $\epsilon>0$ if $0<|x-1|<\frac{\epsilon}{5}$ then $|5x-3-2|<\epsilon,$

which says that $$\lim_{x\rightarrow1}(5x-3)=2$$ by the definition of the limit.

Answer 2

What is the definition of limits ?

$\forall\: \epsilon >0, \exists\: \delta >0 \ni 0<|x-c|<\delta\implies|f(x)-L|<\epsilon$. We then say $\lim_{x\to c} f(x)=L$

In our case $f(x)=5x-3,c=1$ and $L=2$. So let us verify. Given $\epsilon>0$ let there be $\delta=\delta(\epsilon)>0$ such that $0<|x-1|<\delta$ holds.

$\therefore |5x-3-2|=|5x-5|=5|x-1|<\epsilon$. This is actually what we want. But this is easily achieved if we choose $\delta=\frac{\epsilon}{5}$.