How to calculate the problems of limit calculus?

The limit is one of the main branches of calculus that is used to calculate the numerical value of the function and to define the other branches of calculus. The limit is frequently used to calculate the differential of the functions and the numerical value of the functions.

It is also used to define the numerical definite integral of the function by applying the upper and lower limit of the functions. In this lesson, we will cover all the basic intent of limit calculus such as its definition and calculating the problems. 

What is the limit in calculus?

A numerical term that allows the given function to go to some numeric value as the function approach some specific point is known as the limit calculus. The independent variable of the function is very essential for calculating the numerical value of the function. 

If the independent variable whose specific point is given is not present in the function, then the function remains unchanged. The limit calculus is also defined as a function p(w) approached to s is equal to L. Here is the general expression of the limit in calculus.

Lim_w→s p(w) = L

Where w is the independent variable of the function, s is the specific point of the function, p(w) is the given function, and L is the numerical result of the function. 

A limit calculator is a helpful source to find the numerical values of the function according to the above formula of limit calculus.

Types of limits in calculus

There are three basic sub-types of limit in calculus such as,

• Left-hand limit
• Right-hand limit 
• Two-sided limit

Limit Rules in calculus

Here are some basic limit rules in calculus that are used to solve the numerical values of the function. 

Rule Name	Rule
Sum Rule	Limw→s [p(w) + q(w)] = Limw→s [p(w)] + Limw→s [q(w)]
Difference Rule	Limw→s [p(w) – q(w)] = Limw→s [p(w)] – Limw→s [q(w)]
Constant Rule	Limw→s [k] = k
Constant function Rule	Limw→s [k * p(w)] = k * Limw→s [p(w)]
Power Rule	Limt→s [f(t)]2 = [Limt→s [f(t)]2
Product Rule	Limw→s [p(w) * q(w)] = Limw→s [p(w)] * Limw→s [q(w)]
Quotient Rule	Limw→s [p(w) / q(w)] = Limw→s [p(w)] / Limw→s [q(w)]
L’hopital’s Rule	Limw→s [p(w) / q(w)] = Limw→s [d/dw p(w) / d/dw q(w)]

How to calculate the problems of limit calculus?

The limit rules in calculus are frequently used to solve the problems of limit calculus. Here are a few examples of limit calculus solved by using the rules of limit. 

The first example is for calculating the limit problems with the help of the sum and the difference rules of limit calculus. The second example is for calculating the problems of the limit with the help of L’hopital’s rule and the third one is solved by using the product and quotient rule.

Example 1

Find the numerical value of p(w) = 2w³ + 6w⁴ – 7w² + 5 if the particular value is 2.

Solution 

Step 1: First of all, take the given function p(w) and apply the limit notation to it.

p(w) = 2w³ + 6w⁴ – 7w² + 5  

Lim_w→s p(w) = Lim_w→2 [2w³ + 6w⁴ – 7w² + 5]

Step 2: Now apply the limit notation to each function separately with the help of sum and difference laws of limit calculus.

Lim_w→2 [2w³ + 6w⁴ – 7w² + 5] = Lim_w→2 [2w³] + Lim_w→2 [6w⁴] – Lim_w→2 [7w²] + Lim_w→2 [5]

Lim_w→2 [2w³ + 6w⁴ – 7w² + 5] = 2Lim_w→2 [w³] + 6Lim_w→2 [w⁴] – 7Lim_w→2 [w²] + Lim_w→2 [5]

Step 3: Now substitute w = 2 to the above expression. 

= 2 [2³] + 6 [2⁴] – 7 [2²] + [5]

= 2 [2 x 2 x 2] + 6 [2 x 2 x 2 x 2] – 7 [2 x 2] + [5]

= 2 [8] + 6 [16] – 7 [4] + [5]

= 16 + 96 – 28 + 5

= 112 – 28 + 5

= 84 + 5

= 89

Example 2: 

Find the numerical value of p(x) = x⁴ – 16 / (4x² – 4x – 8) with the help of L’hopital’s rule if the particular value is 1.

Solution 

Step 1: First of all, take the given function p(x) and apply the limit notation to it. 

p(x) = x⁴ – 16 / (4x² + 4x – 8)  

Lim_x→s p(x) = Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)]

Step 2: Now apply the limit notation to each function separately with the help of the sum and difference laws of limit calculus.

Lim_x→2 [x⁴ – 16 / (4x² + 4x – 8)] = Lim_x→2 [x⁴] – Lim_x→2 [16] / (Lim_x→2 [4x²] – Lim_x→2 [4x] – Lim_x→2 [8)]

Step 3: Now substitute x = 2 to the above expression. 

Lim_x→2 [x⁴ – 16 / (4x² + 4x – 8)] = [2⁴] – [16] / ([4(2)²] – [4(2)] – 8)]

Lim_x→2 [x⁴ – 16 / (4x² + 4x – 8)] = 16 – 16 / ([4(4)] – [4(2)] – 8)]

Lim_x→2 [x⁴ – 16 / (4x² + 4x – 8)] = 0 / (16 – 8 – 8)]

Lim_x→2 [x⁴ – 16 / (4x² + 4x – 8)] = 0/0

Step 4: As the given function forms the 0/0 form so we have to apply L’hopital’s rule of limit. 

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = Lim_x→2 [d/dx (x⁴ – 16) / d/dx (4x² – 4x – 8)]

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = Lim_x→2 [(4x³ – 0) / (8x – 4 – 0)]

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = Lim_x→2 [(4x³) / (8x – 4)]

Now substitute x = 2

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = [(4(2)³) / (8(2) – 4)]

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = [(4(8)) / (8(2) – 4)]

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = [(32) / (16 – 4)]

Lim_x→2 [x⁴ – 16 / (4x² – 4x – 8)] = [32 / 12] = 8/3 = 2.67

Example 3: For product and quotient rule

Find the numerical value of p(r) = 6r⁴ * 12r⁵ / 11r² * 12 if the particular value is 1.

Solution 

Step 1: First of all, take the given function p(r) and apply the limit notation to it.

p(r) = 6r⁴ * 12r⁵ / 11r² * 12 

Lim_r→a p(r) = Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12]

Step 2: Now apply the limit notation to each function separately with the help of product and quotient laws of limit calculus.

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = Lim_r→1 [6r⁴] * Lim_r→1 [12r⁵] / Lim_r→1 [11r²] * Lim_r→1 [12]

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 6Lim_r→1 [r⁴] * 12Lim_r→1 [r⁵] / 11Lim_r→1 [r²] * Lim_r→1 [12]

Step 3: Now substitute r = 1 to the above expression. 

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 6 [1⁴] * 12 [1⁵] / 11 [1²] * [12]

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 6 [1] * 12 [1] / 11 [1] * [12]

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 6 * 12 / 11 * 12

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 72 / 11 * 12

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 6.55 * 12

Lim_r→1 [6r⁴ * 12r⁵ / 11r² * 12] = 78.6

Wrap up

In this lesson, we have covered all the basics of limit calculus as we have discussed definitions, formulas, and solved examples. Now you can grab all the basics of limit in calculus by learning this post.