# 12.3 Content - Limits

## LIMITS - The Basics

Series and Sequences (Unit 12B) Learning Targets
• Find the limit of a function numerically, graphically and algebraically
• Find and use one-sided limits to determine if two-sided limits exist
• Find the limits of functions and sequences at infinity
• Determine if a function is continuous using the definition of continuity
• Find the rate of change and relate it to something we know

Google defines a limit as "A point or level beyond which something does not or may not extend or pass." So, therefore, a limit is exactly what its name describes - a point at which the numbers can't pass.

Take the limit notation for example:

This says: "the limit, as x approaches c in the equation f(x) from both sides, will approach L, a y-value".

Given a graph, this is easier seen than done.

So, if we were to be given the problem $\lim_{x\rightarrow 6} f(x)$, we need to find what y goes to as x goes to 6.
In this case, y will go toward 4.5. Therefore, $\lim_{x\rightarrow 6} f(x)$ = 4.5

Limits can be found graphically and algebraically. THAT was a graphic approach to finding a limit, we can also find limits algebraically.

Please refer to the bottom of this page to see what limits can mean graphically (also known as graphical significance of limits).

Limits are like the rational functions and their limits we did in first semester.
If you get stuck in the algebra, graph the equation out- you should be able to get a general view of what the limit does.

This is a really cool video that explains the basics of limits!
http://www.calculus-help.com/phobedemo/

# Finding Limits Algebraically

When finding limits algebraically, the easiest thing to do is use the direct substitution method. It works because the X value, when plugged in, is the highest possible value that the Y value could be.
So, just plug in the limit into the equation and you will have your limit.

## Example 1

For this equation, as X approaches 1, then Y approaches 3.

Unfortunately, sometimes the direct substitution method doesn't produce the result that we need. If there is an asymptote or a PORD, then the limit is seen as not existing (DNE).
Since our definition is INFINITELY CLOSE to a number, but not exactly on it, there may still be a limit that works.

Take this example here:

In this equation, if we substitute 0 for X, we are forced to divide by ‍‍‍0- NOT POSSIBLE!‍‍‍
If this happens, do anything in your power to remove the "problem".
In this equation, factoring the numerator for x helps:
We can now cancel x in the numerator and denominator. When we cancel BOTH variables x, we are left with:

Now, we can substitute 0 in for x in 6x - 7 to find the limit is -7:

## ‍Example 2‍

$external image latex.php?latex=\displaystyle\lim_{x+\to+\pm+2}h(x)&bg=ffffff&fg=1c1c1c&s=0$ if $external image latex.php?latex=h(x)=\frac{x+2}{x^2-4}&bg=ffffff&fg=1c1c1c&s=0$.
h(x) is undefined at -2 and 2, so the function needs to be defined at these points. The denominator of h(x) can be factored to $external image latex.php?latex=(x-2)(x+2)&bg=ffffff&fg=1c1c1c&s=0$, thus only leaving $external image latex.php?latex=\displaystyle\lim_{x+\to+-2}h(x)=\lim_{x+\to+-2}\frac{1}{x-2}=-\frac{1}{4}&bg=ffffff&fg=1c1c1c&s=0$

Indeterminant Form = when you plug in the number (or infinity) that the limit is approaching at into the function and get either 0/0, infinity/infinity, or infinity - infinity, then you need to use a different method to find where the limit approaches that number (or infinity); try factoring or multiplying by the conjugate if there are square roots involved.

Continuity and One-sided limits

(This is g(x).)

Given this graph, we can instantly tell one thing: the limit at 1 "appears" to be two numbers.
This is false. The limit cannot come to 2 numbers.
This graph is discontinuous.

So, though $\lim_{x\rightarrow 1} g(x)$
does not exist, we can find what each side goes to as a limit.

Basic notation: - is negative, and + is positive, meaning that it comes from the negative side and the positive side.

So, for $\lim_{x\rightarrow 1-} g(x)$, the y approaches 1, but for $\lim_{x\rightarrow 1+} g(x)$, the y approaches 4. Again, overall $\lim_{x\rightarrow 1} g(x)$= DNE.

Continuity

As stated before, this graph, g(x), is non-continuous. But f(x) way up there is.
In order to determine what an error of a limit means, it's important to see if the graph is continuous or not.
This check is easy.
If Part 1 is equal to Part 2 at the limit given, then it is a continuous graph.

Example

$\left\{\begin{matrix} -log(x) \rightarrow X< 1 \\ x + 1 \rightarrow X \geq 1 \end{matrix}\right.$

So, given the above equation, if‍‍-log(1) = 1+1‍‍, then it is continuous. If not, then it is not continuous.
‍‍
In this particular equation, it is not continuous, because it ends as 0=2, which is false.
This signifies a jump when it is not continuous.
The first equation ends at 0 when x=1, while the second equation starts at 2 when x=1.

Using one sided limits to determine if a limit exists

The best way to determine if a limit exists is by checking the left and right side limits. As previously stated, ‍‍x --> 1-‍‍ means as x approaches 1 from the left or negative side, and x --> 1+ means as x approaches 1 from the right or positive side. The limit can only exist if the number that the left and right side limits each approach is ALL of the following:
• The same number
• A real number
• A finite number

Therefore, if the limit is infinite, imaginary (or complex), or a different number on each side, then the limit Does Not Exist (DNE).

Rate of change

Color key: Blue = function ... Green = rate of change ... Red = points on graph ... Orange = change

Let’s say that the X-axis is effort, and the Y-axis is result, so there is this result: f(x). When you put in x amount of effort, you get f(x) amount of result, and then there is more effort put in: h. There is now x amount of effort put in; however, now that the more effort put in, there is more result as well, so the result f(x) is now more result: f(x+h). At what rate did this change in effort and result happen? To find that, find the change in the result, which is the final result (f(x+h)) minus the initial result (f(x)), and the change in the effort, which is the final effort (x+h) minus the initial effort (x), and divide the change in result by the change in effort, and that gives us the Rate of Change, which looks like this:
$$\frac{f(x+h)-f(x)}{(x+h)-x}$$
Also, do remember the following equation back in algebra class?
$$\frac{y2-y1}{x2-x1 }$$
It is the slope equation. Do you notice how it looks very similar to the rate of change? This is because, wait for it, Rate of Change = Slope!

Problem From Notes - Rate of Change

$$Given f(x) = x^2 - 7x$ $Evaluate \lim_{x\rightarrow 0}\frac{f(x+h) - f(x)}{h}$$
http://www.screenchomp.com/t/qKI82tfQX

Limits at Infinity

When working with x-->infinity, the only thing that we can assume is true is that 1/infinity goes to zero.
So, divide all the terms by the x with the highest exponent.
‍‍‍‍

Extra Problems:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfdirectory/LimitInfinity.html

# Graphical Significance of Limits

## P.O.R.D

$\lim_{x\rightarrow a^{-}}f(x)=f(a)$
$\lim_{x\rightarrow a^{+}}f(x)=f(a)$
$\lim_{x\rightarrow a}f(x)=f(a)$

### Example

$\lim_{x\rightarrow 1}\frac{x^{2}+2x-3}{x^{2}-1}$

## Odd VA

$\lim_{x\rightarrow a^{-}}f(x)=\infty$
$\lim_{x\rightarrow a^{+}}f(x)=-\infty$
$\lim_{x\rightarrow a}f(x)= DNE$

### Example

$\lim_{x\rightarrow -2}\frac{3x^{2}+x-6}{(x+2)(x-4)}$

## Even VA

$\lim_{x\rightarrow a^{-}}f(x)=\infty$
$\lim_{x\rightarrow a^{+}}f(x)=\infty$
$\lim_{x\rightarrow a}f(x)= DNE$

### Example

$\lim_{x\rightarrow -2}\frac{3x^{2}+x-6}{(x+2)^{2}(x-4)}$

## Jump

$\lim_{x\rightarrow a^{-}}f(x)=j$
$\lim_{x\rightarrow a^{+}}f(x)=t$
$\lim_{x\rightarrow a}f(x)= DNE$

### Example

$f(x)\in\begin{Bmatrix} -x+3; & x<0 \\ 3x+1; & x>0 \end{Bmatrix}$

## Endpoint

$\lim_{x\rightarrow a^{-}}f(x)= DNE$
$\lim_{x\rightarrow a^{+}}f(x)=f(a)$
$\lim_{x\rightarrow a}f(x)= DNE$

### Example

$\lim_{x\rightarrow -1}\sqrt{x+1}$

Primary authors of this page (as of 06/02/12):
Moazezi, Sirota, LI, Sherman, Mummadi