## How Do You Compute a Limit Using a Table?

In each of the next two examples, the value of the limit is the same as the value of the function at the point it approaches. This is typically the case for any polynomial.

For functions that are not polynomials, a table i often in order to evaluate the limit. In each of the following two examples, the output of the function grow more positive or more negative. This means the different limits do not exist.

However, a very similar looking fraction may also lead to a limit that does exist. In the next two examples, the limits do exist even though the functions are undefined at the point the x value is approaching.

## How Do You Compute a Limit From a Graph?

In the next several examples, a graph is used to evaluate one-sided limits from the right and left, the corresponding two-sided limit, and the value of the function. The key is to find the y values as the x values approach some value from the left and the right…if those match, the two sided limit is the y value that it matches at.

Does anyone see any mistakes in the examples above?

## How Do You Calculate a Limit Algebraically?

You can recognize the limits by what happens when you substitute the value x approaches into the expression. If it gives 0/0, there is algebra that you can do to find the exact value of the limit.

In the first two examples, the expression may be factored and simplified…then you can substitute the value for x.

$latex \underset{x\to 3}{\mathop{lim }}\frac{{{x}^{2}}-5x+6}{x-3}$

$latex \underset{x\to -1}{\mathop{lim }}\frac{{{x}^{2}}-x-2}{x+1}$

In the next two examples, the fractions in the numerator must be combined before the fraction may be simplified.

$latex \underset{x\to 0}{\mathop{lim }}\frac{\frac{1}{x-6}+\frac{1}{6}}{x}$

$latex \underset{x\to 0}{\mathop{lim }}\frac{\frac{1}{4}-\frac{1}{x+4}}{x}$

The next two examples are designed to throw you off. When you substitute the value into the expression, you do not get 0/0. This means you need to use a table or graph to get the limit.

$latex \underset{x\to -5}{\mathop{lim }}\frac{1}{{{\left( x+5 \right)}^{2}}}$

$latex \underset{x\to 1}{\mathop{lim }}\frac{x}{{{\left( x-1 \right)}^{2}}}$

The next two examples show how to rationalize the numerator to do a limit.

$latex \underset{x\to 1}{\mathop{lim }}\frac{\sqrt{x}-1}{x-1}$

This example may be done two different ways as the next two boards demonstrate.

$latex \underset{x\to 4}{\mathop{lim }}\frac{\sqrt{x}-2}{x-4}$

## How Do You Compute the Limit of a Difference Quotient?

The last part of Section 10.3 asks you to compute several different difference quotients. Some of the problems ask you to compute

$latex \underset{h \to 0}{\mathop{lim }},\frac{f(a+h)-f(a)}{h}$

where f(x) and a are given to you in the problem. Here are a few examples from the board.

$latex f(x)=4x+3$ and $latex a=1$

The board above contains a mistake…do you see where this group made a mistake?

$latex f(x)={{x}^{2}}-4$ and $latex a=1$

$latex f(x)={{x}^{2}}-1$ and $latex a=2$

In the examples below, you are asked to compute a difference quotient containing x instead of a.

Compute $latex \underset{h \to 0}{\mathop{lim }},\frac{f(x+h)-f(x)}{h}$ where $latex f(x)={{x}^{2}}+2x$

Compute $latex \underset{h \to 0}{\mathop{lim }},\frac{f(x+h)-f(x)}{h}$ where $latex f(x)={{x}^{2}}-x$

By the way, the correct solution to the first problem is below.

In this original calculation, f(a+h) and f(a) where switched.

## How Do You Calculate a Limit at Infinity?

You may have notice that there are two ways to use algebra to compute limits at infinity. Let’s look at two possible strategies for computing a limit at infinity for a rational function, $latex \underset{x \to – \infty }{\mathop{lim }},\frac{2{{x}^{3}}+x}{3{{x}^{2}}+1}$.

In the board above, I divided each term in the rational expression by x to the highest power that appears anywhere in the expression. The denominator get very small as the top approaches 2, so the entire expression must grow very big. Since the bottom is also negative, the expression grows big and is negative. Now let’s change that strategy a bit.

In the board above, I divide by x to the largest power that appears in the denominator. The bottom then gets closer and closer to 3. But the top contains 2x which grow more and more negative as x approaches $latex – \infty$. So the entire expression grows more and more negative.

Either strategy gives the correct answer as long as you interpret what is going on in the numerator and denominator correctly.