- 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

Limits can be added and subtracted, but only when those limits exist.

If *a* is a real number and both

and

exist, then

In words, as long as the limits that are added both exist, "the limit of the sum is the sum of the limits.''

= 3 + 9 = 12.

By the previous rule about "pulling out'' constants,

= 4(3) = 12.

Therefore,

= 12 + 9 = 21.

Often we use this rule in reverse. Instead of evaluating

by breaking it up into two limits to evaluate, we would evaluate

by combining the limits into one.

This stuff works for subtraction, too.

If *a* is a real number and both

and

exist, then

In words, as long as the limits that are subtracted exist, "the limit of the difference is the difference of the limits.''

Now, here's a non-example to show why we need each of the limits that are added or subtracted to exist.

Since

does not exist, we are *not* allowed to say

=

We can't evaluate the left-hand side of that equation, but we can evaluate the right-hand side.

If we want to evaluate

we're out of luck.

does not exist,

and we can't subtract "does not exist'' from "does not exist,'' since that doesn't make sense.

Infinity isn't a number, so "∞-∞"' doesn't make sense.

The point is that we can't perform the subtraction

,

since evaluating the limits doesn't give us actual numbers to subtract.

Now for the right-hand side. Since

\frac1x-\frac1x = 0,

we can easily evaluate this limit:

The addition and subtraction rules also extend to adding several functions. as long as the limits

, ,

all exist,

Example 1

Given that \lim_x\to2 find \lim_x\to23 |

Example 2

Assuming the limit exists, find \lim_x\to3 given that \lim_x\to3( |

Example 3

Find \lim_x\to5 given that this limit exists and \lim_x\to5(x- |

Example 4

Assuming that \lim_x\to 10 \lim_x\to 10(4x + 2-3 find \lim_x\to 10 |

Exercise 1

Given that

\lim_x\to12*f*(*x*) = 7,

evaluate the following limit.

Exercise 2

Given that

\lim_x\to12*f*(*x*) = 7,

evaluate the following limit.

Exercise 3

Given that

\lim_x\to12*f*(*x*) = 7,

evaluate the following limit.

\lim_x\to12(10-2*f*(*x*))

Exercise 4

Given that

\lim_x\to12*f*(*x*) = 7,

evaluate the following limit.

\lim_x\to12(*f*(*x*) + 3x-1)

Exercise 5

Given that

\lim_x\to12*f*(*x*) = 7,

evaluate the following limit.

\lim_x\to1210*f*(*x*)-\lim_x\to12(3 + 5*f*(*x*))

Exercise 6

Assuming that \lim_x\to4*f*(*x*) exists, find this limit if

\lim_x\to420*f*(*x*) = 100

Exercise 7

Assuming that \lim_x\to4*f*(*x*) exists, find this limit if

\lim_x\to4(5*f*(*x*) + 11) = 26

Exercise 8

Assuming that \lim_x\to4*f*(*x*) exists, find this limit if

\lim_x\to4(6x-2*f*(*x*)) = 26

Exercise 9

Assuming that \lim_x\to4*f*(*x*) exists, find this limit if

\lim_x\to4(2-3*x* + 10*f*(*x*)) = 0

Exercise 10

Assuming that \lim_x\to4*f*(*x*) exists, find this limit if

\lim_x\to4(2*f*(*x*) + x)-\lim_x\to4(*f*(*x*)-x) = 10