Introduction to Rules of Limits in Calculus Explained with Examples

July 14, 2022

In calculus, the rules of limits are very essential for solving complex problems accurately or figuring out how examples were solved. Limits are a way to find the numerical value of the function by applying the specific point in the place of the corresponding variable.

Limits are very helpful for defining integral calculus & derivative calculus. It is also very essential in the continuity, Taylor series, & Maclaurin series. In this article, we will study the rules of limits with solved examples.

What is the Limit in Calculus?

In mathematics, a limit is a basic type of calculus that finds a value that tells about a function that approaches some value as the input of that function gets closer & closer to some particular point. The equation used to denote the limits of the function is:

p(v) = X

“v” is the corresponding variable of the function at which you have to substitute the particular point
p(v) is the function with variable “v”
“n” is the specific point that “v” approaches.
The numerical output after placing the limit value is “X”

Rules of Limits in Calculus with Examples

In calculus, problems of limits can be solved either by using its rules or a limit solver. There are some basic rules of limits used to solve the complex problems of calculus. Let us discuss the rules of limits briefly with examples.

1. Power and Constant Rules of Limits in Calculus with Examples

The power rule of limits deals with the exponents of the functions. According to this rule, the power of function can be taken after applying the limit value.

[p^m(v)] = [p(v)]^m

While the constant rule of limits tells that the limit value is only applied in a corresponding variable function. The constant functions remain unchanged after applying the limit values.

[A] = A, where A is any constant.

2. Sum Rules of Limits in Calculus with Examples

According to this rule of limits, the notation of limits is applied to each function separately with an addition sign among them. The general equation of this rule of limits in calculus is:

[p(v) + q(v)] = [p(v)] + [q(v)]
[p(v) + q(v) + r(v)] = [p(v)] + [q(v)] + [r(v)]

And so on.

Example

Evaluate 3v²+ 2v as “v” approaches to 2.

Solution

Step 1: Write the given function according to the sum rules with the notation of limits.

[3v²+ 2v] = [3v²)] + [2v]

Step 2: Substitute 2 in the place of “v”.

[3v²+ 2v] = [3(2))] + [2(2)]

= [3(4)] + [2(2)]

→ ‘yoast-text-mark’>= 12 + 4

=’yoast-text-mark’>= 16

3. Multiplication & Division Rules of Limits in Calculus

According to the multiplication & division rules of limits, the limit’s notation is applied to each function separately with cross & division signs among them. The general equation of these rules of limits in calculus are:

[p(v) x q(v)] = [p(v)] x [q(v)]
[p(v) / q(v)] = [p(v)] / [q(v)]
[p(v) x q(v) / r(v)] = [p(v)] x [q(v)] / [r(v)]

And so on.

Example

Evaluate 12v² x 6v / 4 as “v” approaches to 3.

Solution

Step 1: Write the given function according to the multiplication & division rules with the notation of limits.

[12v² x 6uv / 4] = [12v²)] x [6v] /[4]

Step 2: Substitute 3 in the place of “v”.

[12v² x 6uv / 4] = [12(3)²)] x [6(3)] / [4]

= [12(9))] x [6(3)] / [4]

→ ‘yoast-text-mark’>= 108 x 18 / 4

=’yoast-text-mark’>= 1944 / 4

=’yoast-text-mark’>= 486

Use a limit calculator with steps to solve the problems of limits calculus in a fraction of seconds.

4. Difference Rules of Limits in Calculus with Examples

According to this rule of limits, the limit’s notation is applied to each function separately with a subtraction sign among them. The general equation of this rule of limits in calculus is:

[p(v) – q(v)] = [p(v)] – [q(v)]
[p(v) – q(v) – r(v)] = [p(v)] – [q(v)] – [r(v)]

And so on.

Example

Evaluate 5v² – 3v – 4 as “v” approaches to 6.

Solution

Step 1: Write the given function according to the difference rules with the notation of limits.

[5v² – 3v – 4] = [5v²)] – [3v] -[4]

Step 2: Substitute 6 in the place of “v”.

[5v² – 3v – 4] = [5(6)²)] – [3(6)] – [4]

= [5(36))] – [3(6)] – [4]

= [180)] – [18] – [4]

→ ‘yoast-text-mark’>= 162 – 4

=’yoast-text-mark’>= 158

5. L’hopital’s Rule of Limit in Calculus with Examples

L’hopital’s rule of limits is used when a function gives undefined for like 0/0, infinity/infinity/, 0^inf after applying the limit value. According to L’hopital’s rule of limits, you have to take the derivative of the numerator and denominator of the function and substitute the limit value.

If the function again gives an undefined form, repeat the above process again.

[p(v) / q(v)] = [d/dv p(v)] / d/dv q(v)]

Example

Evaluate (12v² – 6v – 6) / (2v² – 2) as “v” approaches to 1.

Solution

Step 1: Write the given function according to the difference & division rules with the notation of limits.

[(12v² – 6v – 6) / (2v² – 2)] = [(12v² – 6v – 6)] / [(2v² – 2)]

= ([12v²] – [6v] – [6]) / ([2v²] – [2])

Step 2: Substitute 3 in the place of “v”.

[(12v² -6v – 6) / (2v² – 2)] = ([12(1)²] – [6(1)] – [6]) / ([2(1)²] – [2])

= ([12(1)] – [6(1)] – [6]) / ([2(1)] – [2])

→ ‘yoast-text-mark’>= (12 – 6 – 6) / (2 – 2)

=’yoast-text-mark’>= (12 – 12) / (2 – 2)

=’yoast-text-mark’>= 0/0

Step 3: Apply L’hopital’s rule of limit, as the function gives 0/0 form.

[(12v² – 6v – 6) / (2v² – 2)] = [d/dv (12v² – 6v – 6) / d/dv (2v² – 2)]

= [(24v – 6(1) – 0) / (4v – 0)]

→ [(24v – 6 – 0) / (4v – 0)]

= [(24v – 6) / (4v)]

Step 4: Substitute 1 again in the place of “v”

[(12v² – 6v – 6) / (2v² – 2)] = [24v] – [6] / [4v]

= [24(1)] – [6] / [4(1)]

= 24 – 6 / 4

→ 18 / 4

= 9 / 2

= 4.5

Summary

In this post, we have learned the basic rules of limits with examples. After reading the above post, you can solve any complex problem of limit calculus easily. You can grab all the basics of this topic from this post.

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