Question

Prove that the limit of $f\left(x\right)=2x-3$ as $x$ approaches $5$ is $7$ using the $ϵ-\delta$ proof.

Answer 1

The function $f\left(x\right)=2x-3$ is a polynomial and as such it is continuous for every $x\in R$. Then, $\underset{x\to 5}{lim}f\left(x\right)=f\left(5\right)=2\times 5-3=10-3=7$

To prove it using the definition of limit,

$|f\left(x\right)-7|=|2x-3-7|=|2x-10|=2|x-5|$

For $x\in (5-f,5+f)$ with $f>0$ then we have $|f\left(x\right)-7|=2|x-5|<2f$

Given any $\in >0$, choose $f_{\in }<\in /2$ s.t

$x\in (5-f_{in},5+f_{\in })$

$\Rightarrow |f\left(x\right)-7|<2f_{\in }<\in$

which proves the limit.