- Topics At a Glance
**Limits**- Functions Are Your Friend
- Graphing and Visualizing Limits
- Piecewise Functions and Limits
- One-Sided Limits
- Limits via Tables
**Limits via Algebra**- All About Asymptotes
- Vertical Asymptotes
- Finding Vertical Asymptotes
- Vertical Asymptotes vs. Holes
- Limits at Infinity
- Natural Numbers
- Limits Approaching Zero
- Estimating a Circle
- The Cantor Set and Fractals
- Limits of Functions at Infinity
- Horizontal, Slant, and Curvilinear Asymptotes
- Horizontal Asymptoes
- Slant Asymptotes
- Curvilinear Asymptotes
- Finding Horizontal/ Slant/ Curvilinear Asymptotes
- How to Draw Rational Functions from Scratch
- Comparing Functions
- Power Functions vs. Polynomials
- Polynomials vs. Logarithmic Functions
- Manipulating Limits
- The Basic Properties
- Multiplication by a Constant
- Adding and Subtracting Limits
- Multiplying and Dividing Limits
- Powers and Roots of Limits
- In the Real World
- I Like Abstract Stuff; Why Should I Care?
- How to Solve a Math Problem

Most of the time, it's more precise to find the limit using algebra.

When finding a limit of the form , the first thing to do is plug *a* into the function to see if it exists. If *f*(*a*) exists, then that's the answer.

Find .

The first thing we do is plug in 3 and see if we find a value.

That's a perfectly good fraction, therefore

Here's an example of how we might not find a solution.

Find .

The first thing we do is plug in 0 and see if we find a number.

= ?

The fraction is undefined, since we can't divide by zero.

Find .

The first thing we do is see if we can plug in 2.

This answer is undefined.

When we're asked to find the limit of a quotient, if we plug in a number and find there's a good chance we can do something about it: we simplify the quotient and try again.

Find .

We already tried to plug in 2 and that didn't work. We'll simplify the fraction by factoring the polynomials.

*x*^{2}-*x*-2 = (*x* + 1)(x-2)

and

*x*^{2}-5*x* + 6 = (x-2)(x-3).

Now we can see why we got before: 2 is a root of the polynomial in the numerator, and also a root of the polynomial in the denominator. Here's the point where we can do something useful: cancel the term (x-2) from the numerator and denominator:

.

Finally, put in 2 again:

We can now say

= 2.

Why does this work? Let

* *

*Now we'll think about .*

*The quotient f(x) can be factored as*

* which equals (x-1) for every value of x except -1 (when x = -1, the quotient is undefined).*

*We'll say that again, because it's important. After factoring f, we see that we can think of it as*

.

*If we graph this, we find the line x-1 with a "hole" in the graph at x = -1 since f(-1) is undefined:*

*It's like the function f(x) is trying to be x-1, but failing at one spot (poor function!). Here's the good news: since f(x) is trying to be x-1, we can find *

*by instead finding*

*.*

Exercise 1

What function do we get if we simplify the quotient?

- \fracx
^{2}+ 6x + 8x^{2}-x-20

Exercise 2

What function do we get if we simplify the quotient?

\item

\fracx^{3}-9xx^{2}-3x

Exercise 3

What function do we get if we simplify the quotient?

- \fracx + 6x
^{2}+ 4x-12

Exercise 4

Find the limits.

- \lim_x\to2\fracx-1x
^{2}-1

Exercise 5

- \lim_x\to1\fracx-1x
^{2}-1

Exercise 6

- \lim_x\to0\fracx
^{3}-xx^{3}+ 6x^{2}+ 9x

Exercise 7

- \lim_x\to2\fracx
^{3}-4xx^{2}-9*x*+ 14

Exercise 8

- \lim_x\to10\fracx
^{2}-100x^{2}+ 6x-40