# How do you use the epsilon delta definition to find the limit of (2+4x)/3 as x approaches 1?

Mar 21, 2018

First, a slightly technical point - you can't really find a limit using the $\epsilon - \delta$ definition. What you can do is, given a number $L$, prove that it is the limit!

#### Explanation:

It seems obvious that

${\lim}_{x \to 1} \frac{2 + 4 x}{3} = \frac{2 + 4 \times 1}{3} = 2$

Let us prove that 2, indeed, is the limit.

Recall that the $\epsilon - \delta$ definition of the limit states that:

A number $L$ is called the limit of a function $f : \mathbb{R} \to \mathbb{R}$ as $x \to a$ if $\forall \epsilon > 0 , \exists \delta > 0$ such that
$0 < | x - a | < \delta \implies | f \left(x\right) - L | < \epsilon$

Shorn of the Greek, this means that we can keep $f \left(x\right)$ as close to $L$ as we want, by keeping $x$ sufficiently close to $a$.

For the current problem

$f \left(x\right) - L = \frac{2 + 4 x}{3} - 2 = \frac{4}{3} \left(x - 1\right)$

so $| f \left(x\right) - L | < \epsilon \iff \frac{4}{3} | x - 1 | < \epsilon$

So, for any given positive $\epsilon$, choose

$0 < \delta < \frac{3}{4} \epsilon$

Then, $| x - 1 | < \delta \implies | x - 1 | < \frac{3}{4} \epsilon \implies | f \left(x\right) - L | = \frac{4}{3} | x - 1 | < \epsilon$

Hence such a $\delta$ exists for every positive $\epsilon$, and thus 2 is the limit!