By Martha Simons


AP Calculus: How to solve limits algebraically

AP Calculus: How to solve limits algebraically

Limits are a vital part of Calculus either you are taking a course or a full-fledged degree in an institution. The topic of limits has evolved with the advancement of Calculus and now its applications in Calculus are wide and significant.

Limits can be used with derivatives, integrations, sequences, and anywhere you want to approach some function with a certain value. Yes, limits make it easy to approach a function at some specific point but, for beginners, it is not easy to solve them.

You may have learned or already know the rules of limits i.e., sum rule, quotient rule, and L’Hopital rule, etc. I will not discuss them here. Instead, I will explain how you can solve limits algebraically. By the way, a limit can also be calculated using graphs as shown in the image below.

solve limit graphic

Let’s understand what limits are before we move on to the algebraic method. So, tighten your seatbelt because it’s time to learn something new.

What are Limits - Formal Definition

As I have already defined limits informally in the previous section, here is the formal definition of limit.

Let f(x) be defined for all x ≠ a over an open interval comprising a. Let L be a real number. Then,

`lim from {x → a} {f left (x right )} = L`

If, for every ε > 0, there exists a δ > 0, such that if 0 < |xa| < δ, then |f(x) − L| < ε.

How can I solve limits algebraically?

Were you aware of the fact that limits can be solved algebraically? There is nothing wrong with solving limits by using algebraic techniques or a limit calculator. The reason is, there is no way one can escape from algebra while studying mathematics.

So, there are three methods from algebra that I will use to solve limits.

  1. Substitution

  2. Factorization

  3. Rationalization

Let’s understand these solving techniques one by one by using a problem from calculus.

  1. Substitution

Substitution is the simplest one in all three methods. This method doesn’t always work well so you should not solely depend on this one. Substitution may also come in handy when using the other two methods.


  • Substitute the limit (value of the variable) in the given equation to approach that specific value.

  • Simplify the equation and get the result of the function.

Let’s solve the below equation if x approaches 3. Substitute the value of x in the equation.

`lim from {x → 3} {{x + 5} over {x − 2}}`

`lim from {x → 3} {{3 + 5} over {3 − 2}}`

= 8/1 = 8

  1. Factorization

If the substitution method doesn’t work for you, then you should try factorization to solve limits. This method is also straightforward as the first one because you may have already learned factorization in school or college.

Here’s what you need to do use this method.


  • Make the factors of numerator and denominator of the equation separately.

  • Cancel the expression in the equation if possible.

  • Get the equation to the simplest form.

  • Substitute the value of limit and simplify.

Let’s solve the limit with factorization.

`lim from {x → 3} {{{x} ^ {2} − 9} over {x − 3}}`

Factorize the numerator.

`lim from {x → 3} {{left (x + 3 right ) left (x − 3 right )} over {x − 3}}`

Cancel, cut, or divide the numerator and denominator if possible.


Now, substitute and simplify.

`lim from {x → 3} {left (3 + 3 right )} = 6`

  1. Rationalization

Rationalization method is a complex one as compared to the previous two methods. Use this method if you spot a square root in the equation.


  • Take the conjugate of the expression.

`lim from {x → 3} {{x − 9} over {sqrt {x} − 3} × {sqrt {x} + 3} over {sqrt {x} + 3}}`

  • Simplify the equation to get the final value.

`lim from {x → 3} {{left (x − 9 right ) left (sqrt {x} + 3 right )} over {sqrt {x} − 3}}`

`lim from {x → 3} {{left (x − 9 right ) left (sqrt {x} + 3 right )} over {x + 3 sqrt {x} − 3 sqrt {x} − 9}}`

`lim from {x → 3} {{left (x − 9 right ) left (sqrt {x} + 3 right )} over {x − 9}}`

Let’s solve the below function with rationalization.

`lim from {x → 3} {{x − 9} over {sqrt {x} − 3}}`

Take the conjugate.

Simplify the above equation.

Substitute the value of x.

`lim from {x → 3} {left (sqrt {x} + 3 right ) = left (sqrt {3} + 3 right ) = 4.73}`

Summing it up

 All of the above-mentioned methods are easily adoptable if you are interested in solving limits with algebraic methods. The limit laws are equally important and these methods cannot replace those rules and laws. Therefore, you should also approach rules of limits when trying to calculate limits.

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Last updated: Sunday, April 19th, 2020 - 12:42PM