# What Are The Types And Rules Of Limit Calculus?

The concept of limit calculus is one of the most essential concepts. Calculus is one of the most important concepts in mathematics, and the limit is one of its most fundamental concepts. Without it, calculus would be much more difficult to understand.

In this article, we’ll provide a brief overview of the limit and outline the types of limits that exist: one-sided, two-sided, etc. We will also provide instructions on how to calculate limits using the rules given.

## What is the limit?

In mathematical terms, a limit is a point at which a function reaches its maximum or minimum value. A limit is simply something that you cannot exceed. It’s used to find out exactly how a function will change as the input values get closer and closer to some given value.

This can be helpful in many different circumstances, such as when trying to figure out how much money someone will make over their lifetime, or when trying to design a machine that can handle more complicated tasks.

The general expression of limit calculus is:

Limu→a f(u) = M

## Types of limits in calculus

There are different types of limits:

1. Pointwise limits

Pointwise limits are the simplest type and they are just what they sound like – the limit is found at one specific point in space. The function of the pointwise limit will be:

1. One-sided limits

One-sided limits involve finding the limit as both the input values and output values approach a single point in space. It can be either a left-hand limit or a right-hand limit.

Limu→a+ f(u) = M

Or

Limu→a f(u) = M

1. Two-sided limits

Two-sided limits involve finding both the pointwise and one-sided limits at different points in space. When both left-hand and right-hand limits exist, we will get the two-sided limits.

Limu→a f(u) = M

1. Limits at infinity

Limits at infinity are important because they tell us how far away from a given point an object or function can get before it stops changing or reaching a new level.

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Limu→ f(u) = M

## Rules of limit calculus

Here are some well-known rules of limit calculus.

1. Constant Rule

The constant rule of limit calculus is used when a function is given in which there is no independent variable available. This rule state that the limit of the constant function remains unchanged.

Limu→a k = k

1. Constant time function rule

This rule is used when there are constant coefficients along with the independent variables of the function. This rule states that the constant coefficient will be written outside the limit notation.

Limu→a kf(u) = k Limu→a f(u)

1. Sum Rule

The sum rule is used when there are two or more terms or functions given along with the plus sign among them. This rule states that the notation of the limit will apply to each function separately.

Limu→a [f(u) + g(u)] = Limu→a [f(u)] + Limu→a [g(u)]

1. Difference Rule

The difference rule is used when there are two or more terms or functions are given along with the minus sign among them. This rule states that the notation of the limit will apply to each function separately.

Limu→a [f(u) – g(u)] = Limu→a [f(u)] – Limu→a [g(u)]

1. Product Rule

The product rule is used when there are two or more terms or functions given along with the multiply sign among them. This rule states that the notation of the limit will apply to each function separately.

Limu→a [f(u) x g(u)] = Limu→a [f(u)] x Limu→a [g(u)]

1. Quotient Rule

The quotient rule is used when there are two or more terms or functions given along with the division sign among them. This rule states that the notation of the limit will apply to each function separately.

Limu→a [f(u) / g(u)] = Limu→a [f(u)] / Limu→a [g(u)]

1. L’hopital’s Rule

L’hopital’s rule is used when a function makes an undefined form such as 0/0, ∞/∞, ∞, etc. This rule states that if the function forms an undefined form, then take the derivative of the given function with respect to the corresponding variable.

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After that apply the limit value again. If the function again forms an undefined form then take the second derivative of the function, and so on until you get the result that is defined.

Limu→a [f(u) / g(u)] = Limu→a [d/du f(u) / d/du g(u)]

To get the step-by-step solution to complex limit problems according to the above rules, try a limit calculator with steps by Allmath.

## Some well-known formulas of limit calculus

Here is a list of some well-known formulas of limit in calculus.

## How to calculate the limit calculus problems?

The problems of limit calculus can be calculated easily with the help of limit rules. Let us take a few examples to understand it.

Example 1

Evaluate the limit of g(w) = 2w2 + 12w4 – 6w6 * 8w2, as the specific point is 3.

Solution

Step 1: First of all, write the given function in the form general limit expression.

Limw→a [g(w)] = Limw→3 [2w2 + 12w4 – 6w6 * 8w2]

Step 2: Now apply the sum, difference, and product rules of limit calculus as there is a plus, minus, and multiply sign among the terms of the function.

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = Limw→3 [2w2] + Limw→3 [12w4] – Limw→3 [6w6] * Limw→3 [8w2]

Step 3: Now use the constant times function rules as there are constant coefficients along with the independent variable.

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = 2Limw→3 [w2] + 12Limw→3 [w4] – 6Limw→3 [w6] * 8Limw→3 [w2]

Step 4: Now apply the specific point to the above expression in place of the independent variable.

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = 2 [32] + 12 [34] – 6 [36] * 8 [32]

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = 2 [9] + 12 [81] – 6 [729] * 8 [9]

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = 18 + 972 – 4374 * 72

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Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = 18 + 972 – 314928

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = 990 – 314928

Limw→3 [2w2 + 12w4 – 6w6 * 8w2] = -313938

Example 2

Evaluate the limit of g(x) = (2x2 – x – 1) / (3x – 3) as the specific point is 1.

Solution

Step 1: First of all, write the given function in the form general limit expression.

Limx→a [g(x)] = Limx→1 [(2x2 – x – 1) / (3x – 3)]

Step 2: Now apply the difference and quotient rules of limit calculus as there is minus and division sign among the terms of the function.

Limx→1 [(2x2 – x – 1) / (3x – 3)] = (Limx→1 [2x2] – Limx→1 [x] – Limx→1 [1]) / (Limx→1 [3x] – Limx→1 [3])

Step 3: Now use the constant times function rules as there are constant coefficients along with the independent variable.

Limx→1 [(2x2 – x – 1) / (3x – 3)] = (2Limx→1 [x2] – Limx→1 [x] – Limx→1 [1]) / (3Limx→1 [x] – Limx→1 [3])

Step 4: Now apply the specific point to the above expression in the place of the independent variable.

Limx→1 [(2x2 – x – 1) / (3x – 3)] = (2[12] – [1] – [1]) / (3 [1] – [3])

Limx→1 [(2x2 – x – 1) / (3x – 3)] = (2[1] – [1] – [1]) / (3 [1] – [3])

Limx→1 [(2x2 – x – 1) / (3x – 3)] = (2 – 1 – 1) / (3 – 3)

Limx→1 [(2x2 – x – 1) / (3x – 3)] = (2 – 2) / (3 – 3)

Limx→1 [(2x2 – x – 1) / (3x – 3)] = 0/0

Step 5: Now use L’hopital’s law of limit calculus as the given function makes an undefined form.

Limx→1 [(2x2 – x – 1) / (3x – 3)] = Limx→1 [d/dx (2x2 – x – 1) / d/dx (3x – 3)]

Limx→1 [(2x2 – x – 1) / (3x – 3)] = Limx→1 [(4x – 1 – 0) / (3 – 0)]

Limx→1 [(2x2 – x – 1) / (3x – 3)] = Limx→1 [(4x – 1) / (3)]

Apply the specific point again.

Limx→1 [(2x2 – x – 1) / (3x – 3)] = [(4(1) – 1) / (3)]

Limx→1 [(2x2 – x – 1) / (3x – 3)] = [(4 – 1) / (3)]

Limx→1 [(2x2 – x – 1) / (3x – 3)] = [(3) / (3)]

Limx→1 [(2x2 – x – 1) / (3x – 3)] = 1

## Sum Up

Now you can grab all the basics of types and rules of limit calculus from this post as we have discussed almost everything about them along with formulas and examples.