Limits are frequently used in calculus to find the numerical values of the function at a specific point. The limit is usually used to define the differential, definite integral, and continuity. The specific point of a limit can be positive, negative, zero, or undefined.

The problems of the limits can be solved by using the rules. The rules of limits are sum, difference, power, constant, product, quotient, and L’hopital’s. L’hopital’s rule of the limit is only applicable when the function forms an indeterminate form after applying the specific point.

In this article, we will learn about the definition and types of the limits along with a lot of examples.

There Are The Details Of This Article

**What are the limits?**

**What are the limits?**

Before discussing the types of the limit, you must be familiar with the definition and formula of the limits. Let us discuss them briefly.

**Definition of the limit**

In calculus, the limit states a value that a function approaches as the input of the given function becomes closer & closer to a specific number. In other words, a numerical value of a function as a corresponding variable approaches a specific number is called the limit of that function.

**The limit formula**

The formula used to find the limits of the function at a specific point with respect to the corresponding variable is given below.

** h(y) = U**

- y is the corresponding variable of the function.
- h(y) is the required function.
- a is the specific point of the function.
- U is the numerical result of the function after applying the limit value.

You have to apply the value of the limit in the corresponding variable, to solve the problems of limits in calculus. The function remains the same if the function is constant or contains other variables. Let us discuss the types of limits with examples.

**Types of limits in calculus**

In calculus, there are three types of limits that are frequently used to solve the problems of the limit.

- The left-hand limit in calculus
- The right-hand limit in calculus
- The two-sided limit in calculus

These types of limits are widely used in calculus for solving complex problems.

**The left-Hand Limit in calculus **

In calculus, the left-hand limit is that type of limit in which the function h(y) approaches a specific value as the variable y approaches a specific number from the left-hand side. The left-hand limits are denoted by the negative power of the specific number.

The formula or equation of the left-hand limit is:

** h(y) = U**

To express the terms that come from the left-hand side and should be less than a, you have to put the negative sign in the equation.

**Example: For the left-hand limit**

Calculate the left-hand limit of 3y^{2} + 3cos(y) + 2y^{3}z – 4, as y approaches to 1.

**Solution **

**Step 1:** Write the given data and apply the notation of left hand limit.

h(y) = 3y^{2} + 3cos(y) + 2y^{3}z – 4

a = 1

** **h(y) = (3y^{2} + 3cos(y) + 2y^{3}z – 4)

**Step 2:** Apply the left-hand limit notation separately to each function by using the sum and difference rules of limits.

(3y^{2} + cos(y) + y^{3}z – 4) = (3y^{2}) + (3cos(y)) + (2y^{3}z) – (4)

**Step 3:** Now write the constant coefficients of the function outside the limit notation by using the constant function rule of the limit.

(3y^{2} + cos(y) + y^{3}z – 4) = 3 (y^{2}) + 3 (cos(y)) + 2z (y^{3}) – (4)

**Step 4:** Now substitute 1 in the place of y to find the limit of the given function by using the power and constant rule of limit.

(3y^{2} + cos(y) + y^{3}z – 4) = 3 (1^{2}) + 3 (cos (1)) + 2z (1^{3}) – (4)

(3y^{2} + cos(y) + y^{3}z – 4) = 3 (1^{ }* 1) + 3 (cos (1)) + 2z (1^{ }* 1 * 1) – (4)

(3y^{2} + cos(y) + y^{3}z – 4) = 3 (1) + 3 (cos (1)) + 2z (1) – (4)

(3y^{2} + cos(y) + y^{3}z – 4) = 3 + 3cos (1) + 2z – 4

(3y^{2} + cos(y) + y^{3}z – 4) = 3cos (1) + 2z – 1

To verify the result, you can use a limit calculator with steps. This calculator will give you step-by-step solutions to the given problems.

**The Right-Hand Limit in calculus **

In calculus, the right-hand limit is that type of limit in which the function h(y) approaches a specific value as the variable y approaches a specific number from the right-hand side. The right-hand limits are denoted by the positive power of the specific number.

The formula or equation of the right-hand limit is:

** h(y) = U**

To express the terms that come from the right-hand side and should be greater than a, you have to put the positive sign in the equation.

**Example: For the right-hand limit **

Calculate the right-limit of 4y^{4} + 6y^{6} + 4xy – 12, as y approaches to 2.

**Solution **

**Step 1:** Write the given data and apply the notation of left hand limit.

h(y) = 4y^{4} + 6y^{6} + 4xy – 12

a = 2

h(y)** **= (4y^{4} + 6y^{6} + 4xy – 12)

**Step 2:** Apply the right-hand limit notation separately to each function by using the sum and difference rules of limits.

h(y)** **= (4y^{4} + 6y^{6} + 4xy – 12) = (4y^{4}) + (6y^{6}) + (4xy) – (12)

**Step 3:** Now write the constant coefficients of the function outside the limit notation by using the constant function rule of the limit.

(4y^{4} + 6y^{6} + 4xy – 12) = 4 (y^{4}) + 6 (y^{6}) + 4x (y) – (12)

**Step 4:** Now substitute 2 in the place of y to find the limit of the given function by using the power and constant rule of limit.

(4y^{4} + 6y^{6} + 4xy – 12) = 4 (2^{4}) + 6 (2^{6}) + 4x (2) – (12)

(4y^{4} + 6y^{6} + 4xy – 12) = 4 (2^{ }* 2 * 2 * 2) + 6 (2^{ }* 2 * 2 * 2 * 2 * 2) + 4x (2) – (12)

(4y^{4} + 6y^{6} + 4xy – 12) = 4 (16) + 6 (64) + 4x (2) – (12)

(4y^{4} + 6y^{6} + 4xy – 12) = 64 + 384 + 8x – 12

(4y^{4} + 6y^{6} + 4xy – 12) = 448 + 8x – 12

(4y^{4} + 6y^{6} + 4xy – 12) = 436 + 8x

**The two-sided Limit in calculus **

When both left and right hand limits exist then the function is said to be the two-sided limit function.

** h(y) = U**

**Example: For the two-sided limit**

Calculate the two-sided limit of 4y^{3} – 6y^{2} + 1, when y approaches 1.

**Solution **

**Step 1:** Write the given data and apply the notation of left hand limit.

h(y) = 4y^{3} – 6y^{2} + 1

a = 1

h(y)** **= (4y^{3} – 6y^{2} + 1)

**Step 2:** Apply the two-sided limit notation separately to each function by using the sum and difference rules of limits.

(4y^{3} – 6y^{2} + 1) = (4y^{3}) – (6y^{2}) + (1)

**Step 3:** Now write the constant coefficients of the function outside the limit notation by using the constant function rule of the limit.

(4y^{3} – 6y^{2} + 1) = 4 (y^{3}) – 6 (y^{2}) + (1)

**Step 4:** Now substitute 1 in the place of y to find the limit of the given function by using the power and constant rule of limit.

(4y^{3} – 6y^{2} + 1) = 4 (1^{3}) – 6 (1^{2}) + (1)

(4y^{3} – 6y^{2} + 1) = 4 (1) – 6 (1) + (1)

(4y^{3} – 6y^{2} + 1) = 4 – 6 + 1

(4y^{3} – 6y^{2} + 1) = -2 + 1

(4y^{3} – 6y^{2} + 1) = -1

**Summary **

Now you can grab all the basics of the types of the limits. In this post, we’ve discussed all the basics of the types of limits. You can solve any problem related to the types of limits easily just by following the above post.

## More Stories

## Benefits of PRALearn

## Photoreading – Terribly Ineffective Speed Reading

## Fundamental Rethinking Of Federal Education Policy