# Limits *(Evaluating)*

*You should read Limits (An Introduction) first*

## Quick Summary of Limits

Sometimes we can't work something out directly ... but we **can** see what it should be as we get closer and closer!

### Example:

\frac{(x^{2} − 1)}{(x − 1)}

Let's work it out for x=1:

\frac{(1^{2 }− 1)}{(1 − 1)} = \frac{(1 − 1)}{(1 − 1)} = \frac{0}{0}

Now 0/0 is a difficulty! We don't really know the value of 0/0 (it is "indeterminate"), so we need another way of answering this.

So instead of trying to work it out for x=1 let's try **approaching** it closer and closer:

### Example Continued:

x | \frac{(x^{2} − 1)}{(x − 1)} | |

0.5 | 1.50000 | |

0.9 | 1.90000 | |

0.99 | 1.99000 | |

0.999 | 1.99900 | |

0.9999 | 1.99990 | |

0.99999 | 1.99999 | |

... | ... |

Now we see that as x gets close to 1, then \frac{(x^{2}−1)}{(x−1)} gets **close to 2**

We are now faced with an interesting situation:

- When x=1 we don't know the answer (it is
**indeterminate**) - But we can see that it is
**going to be 2**

We want to give the answer "2" but can't, so instead mathematicians say exactly what is going on by using the special word "limit"

The **limit** of \frac{(x^{2}−1)}{(x−1)} as x approaches 1 is** 2**

And it is written in symbols as:

*lim***x→1**\frac{x^{2}−1}{x−1} = 2

So it is a special way of saying,* "ignoring what happens when we get there, but as we get closer and closer the answer gets closer and closer to 2"*

As a graph it looks like this: So, in truth, we But we |

## Evaluating Limits

"Evaluating" means to find the value of (*think e-" value"-ating*)

In the example above we said the limit was 2 because it **looked like it was going to be**. But that is not really good enough!

In fact there are **many ways** to get an accurate answer. Let's look at some:

### 1. Just Put The Value In

The first thing to try is just putting the value of the limit in, and see if it works (in other words substitution).

### Example:

limx→10\frac{x}{2} | \frac{10}{2} = 5 |

Easy!

### Example:

limx→1\frac{x^{2}−1}{x−1} | \frac{(1−1)}{(1−1)} = \frac{0}{0} |

No luck. Need to try something else.

### 2. Factors

We can try factoring.

### Example:

*lim***x→1**\frac{x^{2}−1}{x−1}

By factoring (x^{2}−1) into (x−1)(x+1) we get:

*lim*

**x→1**\frac{x^{2}−1}{x−1} =

*lim*

**x→1**\frac{(x−1)(x+1)}{(x−1)}

*lim*

**x→1**(x+1)

Now we can just substitiute x=1 to get the limit:

*lim*

**x→1**(x+1) = 1+1 = 2

### 3. Conjugate

For some fractions multiplying top and bottom by a conjugate can help.

The conjugate is where we change the sign in the middle of 2 terms like this: |

Here is an example where it will help us find a limit:

limx→4 \frac{2−√x}{4−x} | Evaluating this at x=4 gives 0/0, which is not a good answer! |

So, let's try some rearranging:

Multiply top and bottom by the conjugate of the top: | \frac{2−√x}{4−x} × \frac{2+√x}{2+√x} | |

Simplify top using (a+b)(a−b) = a^{2} − b^{2}: | \frac{2^{2} − (√x)^{2}}{(4−x)(2+√x)} | |

Simplify top further: | \frac{4−x}{(4−x)(2+√x)} | |

Cancel (4−x) from top and bottom: | \frac{1}{2+√x} |

So, now we have:

*lim*

**x→4**\frac{2−√x}{4−x} =

*lim*

**x→4**\frac{1}{2+√x} = \frac{1}{2+√4} = \frac{1}{4}

**Done!**

### 4. Infinite Limits and Rational Functions

A Rational Function is one that is the ratio of two polynomials: | f(x) = \frac{P(x)}{Q(x)} | |

For example, here , and P(x) = x^{3 }+ 2x − 1:Q(x) = 6x^{2} | \frac{x^{3} + 2x − 1}{6x^{2}} |

By finding the overall Degree of the Function we can find out whether the function's limit is 0, Infinity, -Infinity, or easily calculated from the coefficients.

Read more at Limits To Infinity.

### 5. L'Hôpital's Rule

L'Hôpital's Rule can help us evaluate limits that at seem to be "indeterminate", suc as \frac{0}{0} and \frac{∞}{∞}.

Read more at L'Hôpital's Rule.

### 6. Formal Method

The formal method sets about proving that we can get **as close as we want** to the answer by making "x" close to "a".

Read more at Limits (Formal Definition)