Basic Concept of Limit in Calculus: Definition, Properties, & Calculations

What is Limit Calculus?

In calculus, a number that a function f(u) reaches as the independent variable “u” of the function reaches a provided specific value is said to be a limit. In simple words, a limit is a value that a function approaches as the independent variable goes closer to a particular point.

Mathematically, the limit of a function can be expressed as:

Lim_u→a f(u) = B

The value of the function can be left-handed, right-handed, or two-sided.

• If the variable of the function approaches to the left of a specific point “a” then the obtained value of the limit is said to be a left-hand limit.
• If the variable of the function approaches to the right of a specific point “a” then the obtained value of the limit is said to be a right-hand limit.
• The two-sided limit of a function occurs when the left-hand limit and right-hand limit become equal.

 

Properties of Limit Calculus

There are several properties of limit calculus that are essential to evaluate the limit values of given functions.

Name of Property	Expression
Property of Addition	Limu→a [f(u) + p(u)] = Limu→a [f(u)] + Limu→a [p(u)]
Property of Power	Limu→a [f(u)]n = [Limu→a f(u)]n
Property of Constant	Limu→a [C] = C
Property of Multiplication	Limu→a [f(u) x p(u)] = Limu→a [f(u)] x Limu→a [p(u)]
Property of Constant Multiple	Limu→a [C * f(u)] = C * Limu→a [f(u)]
Property of Division	Limu→a [f(u) / p(u)] = Limu→a [f(u)] / Limu→a [p(u)]
Property of Subtraction	Limu→a [f(u) – p(u)] = Limu→a [f(u)] – Limu→a [p(u)]

 

How to Calculate the Problems of Limit Calculus?

You can use various techniques for evaluating limit problems such as direct substitution method, factorization, rationalization, or online tools. Below are a few examples to understand how to evaluate limit problems manually.

Example 1:

Evaluate the limit of f(u) = 6u⁴ + 3u⁴ – 2u3 ÷ 6u² x 12u + 16 if “u” approaches “5”.

Solution:

Step 1: First of all, write the given function according to the general expression of limit calculus by placing the notation and specific value of the limit.

Lim_u→a [f(u)] = Lim_u→5 [6u⁴ + 3u⁴ – 2u³ ÷ 6u² x 12u + 16]

Step 2: Now use the properties of limit calculus i.e. properties of addition, subtraction, multiplication, and division to apply the limit notation separately.

Lim_u→5 [6u⁴ + 3u⁴ – 2u³ ÷ 6u² x 12u + 16] = Lim_u→5 [6u⁴] + Lim_u→5 [3u⁴] – Lim_u→5 [2u³] ÷ Lim_u→5 [6u²] x Lim_u→5 [12u] + Lim_u→5 [16]

Step 3: Now use the property of constant Multiple to take the coefficient of the independent variable out of the notation of limit.

Lim_u→5 [6u⁴ + 3u⁴ – 2u³ ÷ 6u² x 12u + 16] = 6Lim_u→5 [u⁴] + 3Lim_u→5 [u⁴] – 2Lim_u→5 [u³] ÷ 6Lim_u→5 [u²] x 12Lim_u→5 [u] + Lim_u→5 [16]

Step 4: Now use the property of power and substitute u = 5 to get the value of the limit.

Lim_u→5 [6u⁴ + 3u⁴ – 2u³ ÷ 6u² x 12u + 16] = 6[Lim_u→5 u]⁴ + 3[Lim_u→5 u]⁴ – 2[Lim_u→5 u]³ ÷ 6[Lim_u→5 u]² x 12[Lim_u→5 u] + Lim_u→5 [16]

Place u = 5

= 6 [5]⁴ + 3 [5]⁴ – 2 [5]³ ÷ 6 [5]² x 12 [5] + [16]

= 6 [625] + 3 [625] – 2 [125] ÷ 6 [25] x 12 [5] + [16]

= 3750 + 1875 – 250 ÷ 150 x 60 + 16

Solve the above expression according to the PEMDAS rule to get an accurate result.

= 3750 + 1875 – 250 ÷ 150 x 60 + 16

= 3750 + 1875 – 1.67 x 60 + 16

= 3750 + 1875 – 100 + 16

= 5625 – 100 + 16

= 5525 + 16

= 5541

Hence,

Lim_u→5 [6u⁴ + 3u⁴ – 2u³ ÷ 6u² x 12u + 16] = 5541

Example 2:

Determine the limit value of p(w) = (w² – 3a – 18) / (w – 6) as “a” approaches 6.

Solution:

Step 1: First of all, write the given function according to the general expression of limit calculus by placing the notation and specific value of the limit.

Lim_w→a [p(w)] = Lim_w→6 [(2w² – 9a – 18) / (w – 6)]

Step 2: Now use the properties of limit calculus i.e. properties of subtraction and division to apply the limit notation separately.

Lim_w→6 [(2w² – 9a – 18) / (w – 6)] = (Lim_w→6 [2w²] – Lim_w→6 [9a] – Lim_w→6 [18]) / (Lim_w→6 [w] – Lim_w→6 [6])

Step 3: Now use the property of constant Multiple to take the coefficient of the independent variable out of the notation of limit.

Lim_w→6 [(2w² – 9a – 18) / (w – 6)] = (2Lim_w→6 [w²] – 9Lim_w→6 [a] – Lim_w→6 [18]) / (Lim_w→6 [w] – Lim_w→6 [6])

Step 4: Now use the property of power and substitute u = 5 to get the value of the limit.

Lim_w→6 [(2w² – 9a – 18) / (w – 6)] = (2[Lim_w→6 w]² – 9Lim_w→6 [a] – Lim_w→6 [18]) / (Lim_w→6 [w] – Lim_w→6 [6])

Place w = 6

= (2 [6]² – 9 [6] – [18]) / ([6] – [6])

= (2 [36] – 9 [6] – [18]) / ([6] – [6])

= (72 – 54 – 18) / ([6] – [6])

= (72 – 72) / (6 – 6)

= 0/0

As given function gives 0/0 which is the indeterminate form of the function. We have to use the L'Hôpital's rule for removing the indeterminate form of function. According to the L’hopital’s rule, you have to take the derivative of the numerator and denominator.

Step 5: Now use L'Hôpital's rule.

[(2w² – 9a – 18) / (w – 6)] = [d/dw (2w² – 9a – 18) / d/dw (w – 6)]

[(2w² – 9a – 18) / (w – 6)] = [d/dw (2w²) – d/dw (9a) – d/dw (18)] / [d/dw (w) – d/dw (6)]

= [4w – 9(1) – (18)] / [(1) – (6)]

= [4w – 9 – 18] / [1 – 6]

= [4w – 27] / [–6]

= -(4w – 27)/6

= (-4w + 27)/6

Step 6: Now apply the limit again

Lim_w→6 [(2w² – 9a – 18) / (w – 6)] = Lim_w→6 [(-4w + 27)/6]

= (Lim_w→6 [-4w] + Lim_w→6 [27])/ Lim_w→6 [6]

= (-4Lim_w→6 [w] + Lim_w→6 [27])/ Lim_w→6 [6]

= (-4 [6] + [27])/ [6]

= (-24 + 27)/6

= (3)/6

= 1/2

The problems of limit calculus can also be solved by using online tools to get quicker results. Here are a few online tools that provides the solution of the limit problems with steps.

1. https://www.allmath.com/limit-calculator.php
2. https://www.limitcalculator.online/
3. https://mathforyou.net/en/online/calculus/limit/

 

Final Words

The term limit in calculus plays a vital role in defining other calculus branches like derivative, integral, continuity, etc. The concept of limit is an essential way to master calculus terms. In this post, we’ve discussed some of the concepts of limits with the help of examples.