﻿ Introduction to limits in calculus & how to solve its problems?

# Introduction to limits in calculus & how to solve its problems?

Limits are frequently used in calculus to find the numerical values of the functions at a specific point. After substituting the specific point, the numerical value can be finite or infinite. It is used to define the derivative calculus & integral calculus.

The complex problems of this type of calculus can be solved easily by using rules & types of limits. The power, product, quotient, & sum rules are frequently used in limits. L’hopital’s rule of limits is used if the value of the limit is undefined after applying the specific point.

In this article, we will study the basics of limits in calculus and how to solve its problems along with a lot of examples.

## What are the limits in calculus?

To find the numerical values of the function at a specific point, the limit in calculus is frequently used. It is an output of the function as the input of the function becomes closer to a specific value.

The general equation used to represent limits in calculus is:

limva p(v) = N

• "lim" is the notation of limits used to represent limits in calculus.
• P(v) is the function.
• "v" is the corresponding variable of limit.
• "a" is the given point.
• N is the output of the function after putting "a".

By using the above equation of limits, you can easily apply the specific point on any function to find the numerical result. Keep one thing in mind, the specific point is only applied to the corresponding variable of the function.

## Types of limits in calculus

Below are three basic types of limits in calculus.

Right-hand limit (R-H-L) in calculus
limva+ p(v) = N

Left-hand limit (L-H-L) in calculus
limva- p(v) = N

Two-sided limit (T-S-L) in calculus
limva p(v) = N

### Rules of limits

Below are a few general rules of limits in calculus.

Name of the rule Rule
Power rule limva p n(v)= [limva[p(v)]n
Sum rule limva [p(v) * q(v)] = limva[p(v)] * limva [q(v)]
Product rule limva [p(v) * q(v)] = limva[p(v)] * limva [q(v)]
Constant rule limva F = F, where F is any constant
Constant multiplier rule limva F * p(v)= Flimva[p(v)], where F is any constant
Difference rule limva [p(v) - q(v)] = limva[p(v)] -limva [q(v)]
Quotient rule limva [p(v) / q(v)] = limva[p(v)] /limva [q(v)]
L'hopital's rule limva [p(v) / q(v)] = limva [p'(v) /q'(v)]

## How to solve the problems of limits in calculus?

The problems of limits can be solved easily either by using its rules or a limit calculator. By using a calculator, you can get the

#### Solution

with steps of any function according to the rules of types of limits.

Follow the below steps to find the Solution to the given problems of limits.

1. First of all, you must have a function, respective variable, & particular point.
2. After identifying the given terms, write them according to the equation of limits.
3. If the problem has multiple functions, use the rules of limits.
4. In the end, put the specific point of limit in the obtained expression to get the result.

Below are a few examples of limits

### (i) For left-hand limit

#### Example

Evaluate p(v) = 9v3 + 12v2 – 11v + 29 as “v” approaches to 3.

#### Solution

Step I:Write the given function, variable, & specific point. And write them according to the equation of limits.

p(v) = 9v3 + 12v2 – 11v + 29

variable of function = v

specific point = a = 3

limva- p(v) = limv3- [9v3 + 12v2 – 11v + 29]

Step II:Now write the limit notation with each function separately by using the difference & sum rules of limits.

limva- p(v) = limv3- [9v3 + 12v2 – 11v + 29]= limv3- [9v3] + limv3- [12v2] – limv3- [11v] + limv3- 

Step III:Use the constant multiplier rule of limits & write the constant coefficients of functions outside the limit notation.

limva- p(v) = limv3- [9v3 + 12v2 – 11v + 29] = 9limv3- [v3] + 12limv3- [v2] – 11limv3- [v] + limv3- 

Step IV:use the constant & power rules of limits & substitute v = 3 in the above expression to find the numerical result.

limv3- [9v3 + 12v2 – 11v + 29] = 9  + 12  – 11  + 

limv3- [9v3 + 12v2 – 11v + 29] = 9 [3 x 3 x 3] + 12 [3 x 3] – 11  + 

limv3- [9v3 + 12v2 – 11v + 29] = 9  + 12  – 11  + 

limv3- [9v3 + 12v2 – 11v + 29] = 243 + 108 – 33 + 29

limv3- [9v3 + 12v2 – 11v + 29] = 351 – 33 + 29

limv3- [9v3 + 12v2 – 11v + 29] = 318 + 29

limv3- [9v3 + 12v2 – 11v + 29] = 347

### (ii) For right-hand limit

#### Example

Evaluate p(v) = 2v5 + 12v2 + 5v - 28 as “v” approaches to 1.

#### Solution

Step I: Write the given function, variable, & specific point. And write them according to the equation of limits.

p(v) = 2v5 + 12v2 + 5v - 28

variable of function = v

specific point = a = 1

limva+ p(v) = limv1+ [2v5 + 12v2 + 5v - 28]

Step II: Now write the limit notation with each function separately by using the difference & sum rules of limits.

limva+ p(v) = limv1+ [2v5 + 12v2 + 5v - 28]= limv1+ [2v5] + limv1+ [12v2] + limv1+ [5v] - limv1+ 

Step III: Use the constant multiplier rule of limits & write the constant coefficients of functions outside the limit notation.

limva+ p(v) = limv1+ [2v5 + 12v2 + 5v - 28] = 2limv1+ [v5] + 12limv1+ [v2] + 5limv1+ [v] - limv1+ 

Step IV: use the constant & power rules of limits & substitute v = 1 in the above expression to find the numerical result.

limv1+ [2v5 + 12v2 + 5v - 28] = 2  + 12  + 5  - 

limv1+ [2v5 + 12v2 + 5v - 28] = 2 [1 x 1 x 1 x 1 x 1] + 12 [1 x 1] + 5  - 

limv1+ [2v5 + 12v2 + 5v - 28] = 2  + 12  + 5  - 

limv1+ [2v5 + 12v2 + 5v - 28] = 2 + 12 + 5 - 28

limv1+ [2v5 + 12v2 + 5v - 28] = 14 + 5 - 28

limv1+ [2v5 + 12v2 + 5v - 28] = 19 - 28

limv1+ [2v5 + 12v2 + 5v - 28] = -9

### (iii) For two-sided limit

#### Example

Evaluate p(v) = 12v3 – 15v2 + 32v + 102, when “v” approaches 2.

#### Solution

Step I:Write the given function, variable, & specific point. And write them according to the equation of limits.

p(v) = 12v3 – 15v2 + 32v + 102

variable of function = v

specific point = a = 2

limva p(v) = limv2 [12v3 – 15v2 + 32v + 102]

Step II:Now write the limit notation with each function separately by using the difference & sum rules of limits.

limva p(v) = limv2 [12v3 – 15v2 + 32v + 102] = limv2 [12v3] – limv2 [15v2] + limv2 [32v] + limv2 

Step III:Use the constant multiplier rule of limits & write the constant coefficients of functions outside the limit notation.

limva p(v) = limv2 [12v3 – 15v2 + 32v + 102] = 12limv2 [v3] – 15 limv2 [v2] + 32 limv2 [v] + limv2 

Step IV:use the constant & power rules of limits & substitute v = 2 in the above expression to find the numerical result.

limv2 [12v3 – 15v2 + 32v + 102] = 12 – 15  + 32  + 

limv2 [12v3 – 15v2 + 32v + 102] = 12[2 x 2 x 2] – 15 [2 x 2] + 32  + 

limv2 [12v3 – 15v2 + 32v + 102] = 12 – 15  + 32  + 

limv2 [12v3 – 15v2 + 32v + 102] = 96 – 60 + 64 +102

limv2 [12v3 – 15v2 + 32v + 102] = 36 + 64 +102

limv2 [12v3 – 15v2 + 32v + 102] = 100 +102

limv2 [12v3 – 15v2 + 32v + 102] = 202

### Summary

In this article, we have discussed the definition, types, rules, & solved examples of limits in calculus. Now after reading the above post, you can grab all the basics of limits in calculus. You can easily solve any complex problem of limits in calculus by following the above post. ## Latest Blog topics

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