**What are limits in calculus and how to calculate it by using rules of limits?**

In mathematics, limits are frequently used in calculus to find the limit values of complex functions. In limits, functions may be linear, polynomial, quadric, or constant. It is used to define various branches of calculus like derivative, antiderivative, and continuity.

In this article, we’ll learn the basics of limit, its rules, and solved examples.

**What are the limits?**

According to Brilliant:

A function at a point c in its domain is the value that the function approaches as its argument approaches to c is known as the limit. In other words, the limit finds the numerical value of the function by taking the limit value of the function.

It is very beneficial for calculating various kinds of calculus. The definite integral use limits as upper and lower from of the interval. Limits also determine the behavior of the function near points of interest.

The general equation of the limits is:

**h(x) = N**

in the above equation, c is the limit value of the function, h(x) is the linear, polynomial, or constant function, and N is the numerical result of the function at c.

**Types of limits**

There are three types of limits in calculus.

- The left-hand limit (-)
- The right-hand limit (+)
- Two-sided limit

**How do calculate limits problems by using the rules of limits?**

The problems of the limits can be solved easily by applying the rules of the limits. First of all, you must be familiar with the rules of limits. Below are some well-known rules of the limit.

Name | Rule |

Constant rule | A = A, where A is any constant |

Constant function rule | AF(x) = A F(x) |

Sum rule | [f(x) + h(x)] = f(x) + h(x) |

Difference rule | [f(x) – h(x)] = f(x) – h(x) |

Product rule | [f(x) * h(x)] = f(x) * h(x) |

Quotient rule | [f(x) / h(x)] = f(x) / h(x) |

You can also use a limit calculator with steps to solve the problems of limits by using its rules.

Now let’s solve some examples of limits to learn how to calculate the problems of limits by using its rules.

**Example 1: For the sum rule**

Find the limit by using the sum rule of 3x^{2} + 2x^{3}, when x approaches 3.

**Solution **

**Step 1:** Apply the limit’s notation on the given function.

f(x) + h(x) = 3x^{2} + 2x^{3}

a = 3

[f(x) + h(x)] = (3x^{2} + 2x^{3})

**Step 2:** Now take the general equation of the sum rule of limits.

[f(x) + h(x)] = f(x) + h(x)

**Step 3:** Now put the values in the above equation and apply the limit value.

(3x^{2} + 2x^{3}) = (3x^{2}) + (2x^{3})

(3x^{2} + 2x^{3}) = 3 (x^{2}) + 2 (x^{3}) (by constant function rule)

(3x^{2} + 2x^{3}) = 3 (3^{2}) + 2 (3^{3})

(3x^{2} + 2x^{3}) = 3 (3^{ }x 3) + 2 (3 x 3 x 3)

(3x^{2} + 2x^{3}) = 3 (9) + 2 (27)

(3x^{2} + 2x^{3}) = 27 + 54

(3x^{2} + 2x^{3}) = 81

**Example 2: For the difference rule**

Find the limit by using the difference rule of 2x^{3} – x^{2}, when x approaches 5.

**Solution **

**Step 1:** Apply the limit’s notation on the given function.

f(x) – h(x) = 2x^{3} – x^{2}

a = 5

[f(x) – h(x)] = (2x^{3} – x^{2})

**Step 2:** Now take the general equation of the difference rule of limits.

[f(x) – h(x)] = f(x) – h(x)

**Step 3:** Now put the values in the above equation and apply the limit value.

(2x^{3} – x^{2}) = (2x^{3}) – (x^{2})

(2x^{3} – x^{2}) = 2 (x^{3}) – (x^{2}) (by constant function rule)

(2x^{3} – x^{2}) = 2 (5^{3}) – (5^{2})

(2x^{3} – x^{2}) = 2 (5^{ }x 5 x 5) – (5 x 5)

(2x^{3} – x^{2}) = 2 (125) – (25)

(2x^{3} – x^{2}) = 250 – 25

(2x^{3} – x^{2}) = 225

**Example 3: For the product rule**

Find the limit by using the product rule of x^{2} * 2x, when x approaches 4.

**Solution **

**Step 1:** Apply the limit’s notation on the given function.

f(x) * h(x) = x^{2} * 2x

a = 4

[f(x) * h(x)] = (x^{2} * 2x)

**Step 2:** Now take the general equation of the product rule of limits.

[f(x) * h(x)] = f(x) * h(x)

**Step 3:** Now put the values in the above equation and apply the limit value.

(x^{2} * 2x) = (x^{2}) * (2x)

(x^{2} * 2x) = (x^{2}) * 2 (x) (by constant function rule)

(x^{2} * 2x) = (4^{2}) * 2 (4)

(x^{2} * 2x) = (4^{ }x 4) * 2 (4)

(x^{2} * 2x) = 16 * 2 (4)

(x^{2} * 2x) = 16 * 8

(x^{2} * 2x) = 128

**Example 4: For the L’hospital rule**

Find the limit by using the quotient and L’hospital rule of 5x^{2} – 20 / 10x – 5x^{2}, when x approaches 2.

**Solution **

**Step 1:** Apply the limit’s notation on the given function.

f(x) / h(x) = 5x^{2} – 20 / 10x – 5x^{2}

a = 2

[f(x) / h(x)] = (5x^{2} – 20 / 10x – 5x^{2})

**Step 2:** Now take the general equation of the quotient rule of limits.

[f(x) / h(x)] = f(x) / h(x)

**Step 3:** Now put the values in the above equation and apply the limit value.

(5x^{2} – 20 / 10x – 5x^{2}) = [5x^{2} – 20] / [10x – 5x^{2}]

(5x^{2} – 20 / 10x – 5x^{2}) = [5(2)^{2} – 20] / [10(2) – 5(2)^{2}]

(5x^{2} – 20 / 10x – 5x^{2}) = [5(4) – 20] / [10(2) – 5(4)]

(5x^{2} – 20 / 10x – 5x^{2}) = [20 – 20] / [20 – 20]

(5x^{2} – 20 / 10x – 5x^{2}) = 0 / 0

**Step 4:** The given function gives the 0/0 form so we have to use the L’hospital rule of limits. According to this rule, take the derivative of the numerator and the denominator and then apply the limit value again.

(5x^{2} – 20 / 10x – 5x^{2}) = [d/dx (5x^{2} – 20)] / [d/dx (10x – 5x^{2})]

(5x^{2} – 20 / 10x – 5x^{2}) = [(10x – 0)] / [(10 – 10x)]

(5x^{2} – 20 / 10x – 5x^{2}) = [(10(2)] / [(10 – 10(2))]

(5x^{2} – 20 / 10x – 5x^{2}) = [20] / [10 – 20]

(5x^{2} – 20 / 10x – 5x^{2}) = 20 / -10

(5x^{2} – 20 / 10x – 5x^{2}) = 2 / -1

(5x^{2} – 20 / 10x – 5x^{2}) = -2

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**Summary **

Limits play a vital role in calculus for solving various kinds of problems. In this article, we’ve learned about the basics of limits in calculus and how to apply limit values by using the rules of limits.

What are limits in calculus and how to calculate it by using rules of limits?