# How do you prove that the limit of 3x+5=35 as x approaches 10 using the precise definition of a limit?

First we need to state this precise definition of a limit of a function $f : X \to \mathbb{R}$:

${\lim}_{x \to {x}_{0}} f \left(x\right) = a \in \mathbb{R} \iff$
$\iff \forall \epsilon > 0 \exists \delta > 0 \forall x \in X : | x - {x}_{0} | < \delta \implies | f \left(x\right) - a | < \epsilon$

From a geometrical point of view, this means that saying this limit takes the value $a$ is the same thing as saying that if $x$ is near ${x}_{0}$ , then $f \left(x\right)$ has to be near $a$, and this has to be valid no matter how small we consider the distance between $f \left(x\right)$ and $a$.

For $f \left(x\right) = 3 x + 5$, the proof that ${\lim}_{x \to 10} f \left(x\right) = 35$ is very simple.

Given $\epsilon > 0$, take $\delta = \frac{\epsilon}{3}$. Then, using the triangle inequality:

$| x - 10 | < \delta \implies | 3 x + 5 - 35 | = | 3 x - 30 | = 3 | x - 10 | < 3 \delta = 3 \frac{\epsilon}{3} = \epsilon \iff | 3 x + 5 - 35 | < \epsilon$

This proof can be generalized to any function $f$ that fulfils the condition:

$\exists k \ge q 0 \forall {x}_{1} , {x}_{2} \in X : | f \left({x}_{2}\right) - f \left({x}_{1}\right) | \le q k | {x}_{2} - {x}_{1} |$

Functions of this kind are caled Lipschitz continuous functions.