# How to Solve Calculus Limit Problems

By Lila S. Kallstrom

Falling a bit behind in your tough calculus class? Never fear, review Calculus limit problems here. Find step by step instructions as well as the theory and graphs related to these problems.

## What are Calculus limit problems?

Solving or evaluating functions in math can be done using direct and synthetic substitution. Students should have experience in evaluating functions which are:

1. constant functions such as:

f(x) = c where c is any real number

f(x) = 7 or y = 7

2. polynomial functions such as:

f(x) = 3x5 – x2 + 6x – 2

Sometimes evaluating a function may lead to an undefined form such as 1/0 or ∞/∞. The concept of limits is to evaluate a function as x approaches a value but never takes on that value.

To solve a limit, see the 4 examples of a limit problems involving direct substitution.

When graphing a solution of an equation in calculus, such as example 1, the graph will pass through the y-value 4/3 when x is the value 1. The line will be a straight line and the graph is said to be continuous at x = 1.

## Limits That Need Simplification

Some limits will need simplification before they can be solved:

If direct substitution yields 0/0, undefined, you have to factor & reduce.

Lim x2 - 4

x →2 x - 2

The limit exists even though the function is undefined. The limit for this example is 4.

To solve an undefined limit, see examples 5 and 6 of limits that need simplification.

If direct substitution yields ∞/∞, undefined, then divide by the highest power.

lim 3x + 1

x→∞ 5x + 6

The limit exists even though the function is undefined. The limit for this example is 3/5.

To solve an undefined limit, see examples 7 and 8 of limits that need simplification.

## Limits That Do Not Exist

Some limits are of a form called a Limit that does not exist. This means that the function f(x) does not approach a single value, a, as x → a and we say that the limit of f(x) as x → a does not exist. There will be a break or a discontinuity in the graph at the point where x = a. The function is discontinuous at x = a. The following limit does not exist as x approaches 2.

Lim x2 + 4

x→2 x - 2

This kind of problem is of the form that will result in 0/0 and usually there is a factor in the numerator and denominator using the value a, that x is approaching.

To solve a limit that has the form 0/0, see example 9 of limits that do not exist.

There are also special cases of limits to solve involving the difference of radicals in the numerator and denominator. Factoring polynomials such as the difference of squares or difference of cubes help to simplify these functions into solvable limits.

I know you don't want to hear this, but practice makes perfect! If you're still having a hard time getting it, solve several different examples and practice identifying all the forms.