# The formal definition of Limit — epsilon-delta

A Formal Definition of Limit:
Let $f(x)$ be defined on a open interval about $x_{0}$, except possibly at $x_{0}$ itself. We say that $f(x)$ approaches the limit L as x approaches $x_{0}$, and write $\lim_{x \rightarrow x_{0}} f(x)=L$,

if for every number $\in > 0$, there exists a corresponding number $\delta >0$, such that for x, $0<|x-x_{0}|<\delta \Longrightarrow |f(x)-L|< \in$.

Note: In the definition, $\delta$ is a function of $\in$, or in other words, $\delta$ depends on $\in$.

This definition is useful to prove important theorems about limits, that is, we can then easily evaluate the limits of complicated functions, whose limits are too cumbersome to evaluate from intuitive first principles.

Think of  the definition as follows: We desire an output tolerance so that we need to find appropriate input given by $\delta$.

Show  that $\lim_{x \rightarrow 1}=2$

Solution:

Let $x_{0}=1$, $f(x)=5x-3$ and $L=2$ in the definition of limit. For any given $\in > 0$, we have to find a number $\delta > 0$ such that if $x \neq 1$. and x is within distance $\delta$ of $x_{0}$, that is, if $0 < |x-1|< \delta$, then $f(x)$ is within distance $in$ of $L=2$, that is, $|f(x)-2|< \in$

We find $\delta$ by working backwards from the $\in$ -inequality. $|(5x-3)-2|= |5x-5|<\in$, that is, $5|x-1|< \in$, that is, $|x-1|< \in/5$.

Important note: The value of $\delta=\in/5$ is not  the only value that will make $0 < |x-1|< \delta$ imply $|5x-5|<\in$. Any smaller positive $\delta$ will do as well. The definition does not ask for a “best” positive $\delta$, just one that will work.

Any questions, comments are most welcome…

More later…

Nalin Pithwa

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