Limit calculus is one of the well-known sub-types of calculus in which different properties and types are used to explore limits of the functions at the specific point. The best use of limit calculus is to define the term differential calculus and integral calculus.
Limits also used to check the continuity of the function to confirm that the given function is continuous or discontinuous at the given point. There are several uses of limits in different branches of education and daily life.
You can get the brief information of limits from this post as we are going to explore types of limit calculus, properties of limit calculus, and solved examples of limit calculus.
Limit Calculus
In mathematics, the limit is the sub-type of calculus that is used to find the numerical value of the function at the particular point. The point of the function will let you known whether the function is defined or undefined.
If the function is defined, then the limit value is will be evaluated by placing the specific point to the function and simplify it. If the function is undefined at the given point, then you have to apply various methods of limit calculus to make it defined.
These methods include L’hopital’s rule, rationalization, factorization, and by parts. You Can use any method of limit calculus to explore the function of limit and find the value by making it defined.
Expression of limit Calculus
The expression of limit calculus could be written as:
Limz→a h(z) = H(a) = M
Where
- z is the independent variable of the function
- a is the particular point of the function
- h(z) is the function
- H(a) is the applied value of specific point to the function
- M is the result of limit calculus
Types of Limit Calculus
There are two well-known types of limit calculus:
- One Sided Limit
- Two Sided Limit
One Sided Limit
One-sided limit is the type of limit calculus that is used to explain the behavior of the function as the given input approaches a particular point from either right or left side. When approaches to a specific point, the one sided limit is used to analyze the function.
This type of limit calculus has function value from one direction only that is from left side or right side. It is denoted by the power of “+” or “-” of the variable. Such as if the function comes from the left then the power of the variable would be negative “-”.
And if the power of the variable is positive “+” then it indicates that the function comes from the right. The expressions for the one-sided limit are:
For Left sided limit
Limz→a– h(z) = H(a) = M
For right sided limit
Limz→a+ h(z) = H(a) = M
Two Sided Limit
Two-sided limit is the type of limit calculus that is used to explain the behavior of the function as the given input approaches a particular point from both the right and left sides. When approaches to a specific point, the two-sided limit is used to analyze the function’s behavior from both sides.
This type of limit calculus is also known as limits from both sides and bilateral limits. The expression for the two-sided limit is:
Limz→a–h(z) = H(a) = M
Properties of Limit Calculus
Here are the well-known properties of limit calculus.
| Properties | Expressions |
| PowerProperty | Limz→a f(z)n = [Limz→a f(z)]n |
| Constant Property | Limz→a f(x) = f(x) |
| Constant Function Property | Limz→a kf(z) = k Limz→a f(z) |
| Sum Property | Limz→a [f(z) + h(z)] = Limz→a [f(z)] + Limz→a [h(z)] |
| Difference Property | Limz→a [f(z) – h(z)] = Limz→a [f(z)] – Limz→a [h(z)] |
| Quotient Property | Limz→a [f(z) / h(z)] = Limz→a [f(z)] / Limz→a [h(z)] |
| L’hopital’ Property | Limz→a [f(z) / h(z)] = Limz→a [d/dz (f(z)) / d/dz (h(z))] |
| Product Property | Limz→a [f(z) * h(z)] = Limz→a [f(z)] * Limz→a [h(z)] |
Examples of Limit Calculus
Here are a few examples of limit calculus:
Example 1
Evaluate the limit of f(z) = 15z2 + 16z4 – 2z * 3z5, as the particular point is 2.
Solution
Step 1: Write the given function of limit calculus with the notation of limit along with specific point.
Limz→a [f(z)] = Limz→2 [15z2 + 16z4 – 2z * 3z5]
Step 2: Now apply the properties of sum, difference, and product to the above expression.
Limz→2 [15z2 + 16z4 – 2z * 3z5] = Limz→2 [15z2] + Limz→2 [16z4] – Limz→2 [2z] * Limz→2 [3z5]
Step 3: Now apply the constant function property of limit calculus.
Limz→2 [15z2 + 16z4 – 2z * 3z5] = 15Limz→2 [z2] + 16Limz→2 [z4] – 2Limz→2 [z] * 3Limz→2 [z5]
Step 4: Now use the power property of limit calculus and substitute the limit value.
= 15 [Limz→2 z]2 + 16 [Limz→2 z]4 – 2 [Limz→2 z]1 * 3 [Limz→2 z]5
= 15 [2]2 + 16 [2]4 – 2 [2]1 * 3 [2]5
= 15 [2 x 2] + 16 [2 x 2 x 2 x 2] – 2 [2] * 3 [2 x 2 x 2 x 2 x 2]
= 15 [4] + 16 [16] – 2 [2] * 3 [32]
= 60 + 256 – 4 * 96
= 60 + 256 – 384
= 316 – 384
= -68
You can also use a limit calculator to evaluate the limit problems in seconds without involving into lengthy and time-consuming calculations.
Example 2
Evaluate the limit of f(v) = (3v3 – 3) / (v2 + 2v – 3) as the specific point is 1.
Solution
Step 1: Write the given function of limit calculus with the notation of limit along with specific point.
Limv→a [f(v)] = Limv→1 [(3v3 – 3) / (v2 + 2v – 3)]
Step 2: Now apply the properties of sum, difference, and division to the above expression.
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = (Limv→1 [3v3] – Limv→1 [3]) / (Limv→1 [v2] + Limv→1 [2v] – Limv→1 [3])
Step 3: Now apply the constant function property of limit calculus.
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = (3Limv→1 [v3] – Limv→1 [3]) / (Limv→1 [v2] + 2Limv→1 [v] – Limv→1 [3])
Step 4: Now use the power property of limit calculus and substitute the limit value.
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = (3 [13] – [3]) / ([12] + 2 [1] – [3])
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = (3 [1] – [3]) / ([1] + 2 [1] – [3])
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = (3 – 3) / (1 + 2 – 3)
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = (3 – 3) / (3 – 3)
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = 0/0
Step 5: Now use L’hopital’s law of limit calculus as the given function makes an indeterminate form.
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = Limv→1 [d/dv (3v3 – 3) / d/dv (v2 + 2v – 3)]
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = Limv→1 [d/dv (3v3) – d/dv (3) / d/dv (v2) + d/dv (2v) – d/dv (3)]
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = Limv→1 [(9v2) – (0) / (2v) + (2) – (0)]
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = Limv→1 [9v2 / (2v + 2)]
Apply the specific point again.
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = [9(1)2 / (2(1) + 2)]
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = [9(1) / (2(1) + 2)]
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = [9 / (2 + 2)]
Limv→1 [(3v3 – 3) / (v2 + 2v – 3)] = [9/4]
Final Words
We have explored the definition, types, and properties, and solved examples of limit calculus. Now you can grab all the basics of this topic from this post as we explained with the step-by-step method to evaluate the limit.
