Limits of a Function

Delta Epsilon Definition of a Function on an Interval

Let $I$ be an open interval containing $a$ , and let $f$ be a function defined on $I$ , except possibly at $a$ . The limit of $f(x)$ as $x$ approaches $a$ is $L$ , donoted as

$\lim_{x \to a} f(x) = L,$

means that given any $\epsilon > 0$ , there exists $\delta > 0$ such that for all $x \neq a$ , if $\abs{x - a} < \delta$ , then $\abs{f(x) - L} < \epsilon.$

This means that when we pick any value of $\epsilon > 0$ first to make a range of $y$ values around $L$ , that is, $\abs{y - L} < \epsilon$ , if we can always find a value of $\delta > 0$ around $a$ such that when $\abs{x - a} < \delta$ , $\abs{f(x) - L} < \epsilon$ , then the limit as $x$ approaches $a$ is $L$ .

Delta Epsilon Limit

Example

Prove

$\lim_{x \to 1} \frac{3x(x-1)}{x-1} = 3.$

We need to show that given $\epsilon > 0$ , there exists $\delta > 0$ such that

$0 < \abs{x - 1} < \delta \implies \abs{\frac{3x(x-1)}{x-1} - 3} < \epsilon$

Basically, we need to find $\delta$ as a function of $\epsilon$ here to show that no matter what the value of $\epsilon$ is, we can find a satisfactory $\delta$ .

We can do this with some algebraic manipulation:

$\abs{\frac{3x(x-1)}{x-1} - 3} < \epsilon \iff \abs{3x - 3} < \epsilon \iff \abs{x -1} < \frac{\epsilon}{3}$

Now we've shown that as long as $\abs{\delta} < \frac{\epsilon}{3}$ , then $\abs{f(1 - \delta) - 3} < \epsilon$ .

Limits of a Function on a Set of Real Numbers

This is more general than the definition above in that it applies to the case where there domain is a set of real numbers, not just an interval.

Let $A$ be a set of real numbers and $f(x)$ be a function $f : A \to \reals.$ The limit of $f(x)$ as $x$ approaches $a$ is $L,$ written

$\lim_{x \to a}{f(x)} = L$

if for every $\epsilon > 0$ there exists some $\delta > 0$ such that $\abs{f(x) - L} < \epsilon$ if $0 < \abs{x - a} < \delta.$ We assume $a$ is a real number and that there are numbers $x \in A$ satisfying $0 < \abs{x - a} < \delta.$

Limits of Function Rules

Suppose $\lim_{x \to a}{f(x)} = L, \lim_{x \to a}{g(x)} = M$ and $c \in \reals.$ Then

$\begin{align*} \lim_{x \to a}{f(x) + cg(x)} & = L + cM, \\ \lim_{x \to a}{f(x)g(x)} & = LM, \\ \lim_{x \to a}{\frac{f(x)}{g(x)}} & = \frac{L}{M}, ~ M \neq 0. \end{align*}$

(same assumption as given above applies for the domains of $f,$ $g$ containing values near $a .$

One-Sided Limits

Let $A$ be a set of real numbers containing $a$ and $f(x)$ be a function $: A \to \reals.$ The limit of $f(x)$ as $x$ approaches $a$ from the right is $L,$ written

$\lim_{x \to a^+}{f(x)} = L$

if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\abs{f(x) - L} < \epsilon$ if $a < x < a + \delta.$ We assume there are numbers $x \in A$ satisfying $a < x < a + \delta.$

Let $A$ be a set of real numbers containing $a$ and $f(x)$ be a function $f : A \to \reals.$ The limit of $f(x)$ as $x$ approaches $a$ from the left is $L,$ written

$\lim_{x \to a^-}{f(x)} = L$

if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\abs{f(x) - L} < \epsilon$ if $a - \delta < x < a.$ We assume there are numbers $x \in A$ satisfying $a - \delta < x < a.$

Theorem: Let $A$ be a set of real numbers containing $a$ and $f(x)$ be $f : A \to \reals,$ and assume that for every $\delta > 0$ there are numbers $x \in A$ satisfying $0 < \abs{x - a} \delta.$ Then $\lim_{x \to a}{f(x)} = L$ if and only if $\lim_{x \to a^-}{f(x)} = \lim_{x \to a^+}{f(x)} = L.$

Limits Involving Infinity

Let $A$ be a set of real numbers and $f(x)$ a function from $A$ into $\reals.$ We say that the limit of $f(x)$ as $x$ approaches infinity is $L,$ written as

$\lim_{x \to \infty}{f(x)} = L,$

if for every $\epsilon > 0$ there exists a real number $M$ such that $\abs{f(x) - L} < \epsilon$ if $x \geq M.$

Let $A$ be a set of real numbers and $f(x)$ a function from $A$ into $\reals.$ We say that the limit of $f(x)$ as $x$ approaches negative infinity is $L,$ written as

$\lim_{x \to - \infty}{f(x)} = L,$

if for every $\epsilon > 0$ there exists a real number $M$ such that $\abs{f(x) - L} < \epsilon$ if $x \leq M.$

Tricks

Factoring a root out of a numerator

Say we have:

$\lim_{n \to \infty} \frac{n + 1}{\sqrt{(n + 1)(2n+1)}}$

Since $n$ is going to infinity it is positive and for positive $n$ , $n + 1 = \sqrt{(n+1)^2}$ we can factor out a root to simplify:

$\lim_{n \to \infty} \frac{\sqrt{(n + 1)^2}}{\sqrt{(n + 1)(2n+1)}} = \lim_{n \to \infty} \sqrt{\frac{(n + 1)^2}{(n + 1)(2n+1)}} = \lim_{n \to \infty} \sqrt{\frac{n+1}{2n+1}}$

From there we can move the limit inside the radical:

$\sqrt{\lim_{n \to \infty} \frac{n + 1}{2n + 1}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$