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

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

means that given any , there exists such that for all $x\ne a$, if , then

This means that when we pick any value of first to make a range of $y$ values around $L$, that is, , if we can always find a value of around $a$ such that when , , then the limit as $x$ approaches $a$ is $L$.

Example

Prove

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

We need to show that given , there exists such that

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

We can do this with some algebraic manipulation:

Now we've shown that as long as , then .

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 $f:A\to \mathbb{R}.$ The limit of $f(x)$ as $x$ approaches $a$ is $L,$ written

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

if for every there exists some such that if We assume $a$ is a real number and that there are numbers $x\in A$ satisfying

Limits of Function Rules

Suppose $\underset{x\to a}{lim}f(x)=L,\underset{x\to a}{lim}g(x)=M$ and $c\in \mathbb{R}.$ Then

(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 $:A\to \mathbb{R}.$ The limit of $f(x)$ as $x$ approaches $a$ from the right is $L,$ written

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

if for every there exists a such that if We assume there are numbers $x\in A$ satisfying

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

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

if for every there exists a such that if We assume there are numbers $x\in A$ satisfying

Theorem: Let $A$ be a set of real numbers containing $a$ and $f(x)$ be $f:A\to \mathbb{R},$ and assume that for every there are numbers $x\in A$ satisfying Then $\underset{x\to a}{lim}f(x)=L$ if and only if $\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)=L.$

Limits Involving Infinity

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

$$\underset{x\to \mathrm{\infty }}{lim}f(x)=L,$$

if for every there exists a real number $M$ such that if $x\ge M.$

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

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=L,$$

if for every there exists a real number $M$ such that if $x\le M.$

Tricks

Factoring a root out of a numerator

Say we have:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{n+1}{\sqrt{(n+1)(2n+1)}}$$

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:

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

From there we can move the limit inside the radical:

$$\sqrt{\underset{n\to \mathrm{\infty }}{lim}\frac{n+1}{2n+1}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$$