What are the limits in calculus? Explained with rules & examples

In mathematics, the limit is frequently used to solve complex calculus problems. It is a method to find the numerical result of the function at a particular point by substituting the value in the corresponding independent variable of the function.

Limits are used to define the continuity, Taylor series, integral calculus, and differential calculus. The definite integral finds the numerical result of the functions with the help of limits. Similarly, differential calculus used limits to find a new function.

 In this article, we will discuss limits definition, rules, and examples.

What are the limits in calculus?

In calculus, a basic term that is used to determine a numerical value that states a function that approaches some result as the given input of that function gets closer to a specific point is said to be the limit.

The general equation used to denote limits along with limit notation is:

lim_t→m s(t) = Y

·         “t” is the independent variable of the function in which you have to put the specific point to get the result.

·         “s(t)” is the given function.

·         The approaching point of “t” is “m”.

·         “Y” is the final result of the function after putting “m” in place of “t”.

Rules of limits in calculus

In calculus, the rules of limits play a vital role in the calculations of various problems. There are several rules of limits in calculus, let us discuss them briefly.

1.   Constant rule of limits

In calculus, when a constant integer or variable is given, then the constant rule of limits is applied. According to the constant rule, the constant functions remain the same after substituting the specific point “m”.

lim_t→m [F] = F, where F is any constant.

Example

Evaluate the limit of 25 as “t” approaches to “6”.

Solution

Step I: Write the given terms according to the limit equation.

lim_t→6 [25]

Step II: Substitute m=6 in the above expression.

lim_t→6 [25] = 25 (by constant rule)

2.   Power rule of limits

In calculus, when an exponential function is given, then the power rule of limits is used. By using this rule, the power of the function is taken after putting the specific point of limits.

The general equation of this rule of limit is:

lim_t→m [s^a(t)] = [lim_t→m s(t)]^a

Example

Evaluate the limit of t⁴ as “t” approaches “2”.

Solution

Step I: Write the given terms according to the limit equation.

lim_t→2 [t⁴] = [[t]]⁴

Step II: Substitute m=2 in the above expression.

lim_t→2 [t⁴] = [lim_t→2 [t]]⁴ = 2⁴ (by power rule)

= 2 x 2 x 2 x 2

= 16

The above problem of limit calculus can also be solved by using a limit solver.

3.   Difference rule of limit in calculus

When a function is given along with a minus sign, then this rule of limits is used. According to the difference rule of limits, the limit notation must be applied to each function separately with a minus sign between the functions.

The general equation of difference rule of limits in calculus is:

·         lim_t→m [s(t) – w(t)] = lim_t→m [s(t)] – lim_t→m [w(t)]

·         lim_t→m [s(t) – w(t) – y(t)] = lim_t→m [s(t)] – lim_t→m [w(t)] – lim_t→m [y(t)]

And so on.

Example

Evaluate 4t – t² as “t” approaches to 4.

Solution

Step I: Write the given terms according to the limit equation.

lim_t→4 [4t – t²]

Step II: Write limit notation with each function separately by using the difference rule of limits.

lim_t→4 [4t – t²] = lim_t→4 [4t] – lim_t→4 [t²]

Step III: Substitute m=4 in the above expression.

lim_t→4 [4t – t²] = [4(4)] – [4²]

lim_t→4 [4t – t²] = [4(4)] – [4 x 4]

lim_t→4 [4t – t²] = 16 – 16

lim_t→4 [4t – t²] = 0

4.   L’hopital’s rule of limit in calculus

When a function forms inf/inf, inf⁰, or 0/0 form after putting the specific point of limits, then L’hopital’s rule of limits is used. According to this rule of limits, the given function must be differentiated with respect to the independent variable and then substitute the value of limits again.

Repeat this process until you get the result. The general equation of this rule is:

lim_t→m [s(t) / w(t)] = lim_t→m [d/dt s(t)] / d/dt w(t)]

Example

Evaluate (4t² – 9t – 9) / (2t² – 18) as “t” approaches to 3.

Solution

Step I: Write the given terms according to the limit equation.

lim_t→3 [(4t² – 9t – 9) / (2t² – 18)]

Step II: Write limit notation with each function separately by using the difference & quotient rules of limits.

lim_t→3 [(4t² – 9t – 9) / (2t² – 18)] = lim_t→3 [(4t² – 9t – 9) /  (2t² – 18)]

lim_t→3 [(4t² – 9t – 9) / (2t² – 18)] = (lim_t→3 [4t²] – lim_t→3 [9t] – lim_t→3 [9]) / (lim_t→3 [2t²] – lim_t→3 [18])

Step III: Substitute m=3 in the above expression.

lim_t→3 [(4t² – 9t – 9) / (2t² – 18)] = ([4(3)²] – [9(3)] – [9]) / ([2(3)²] – [18])

= ([4(9)] – [9(3)] – [9]) / ([2(9)] – [18])

= ([36] – [27] – [9]) / ([18] – [18])

= (36 – 36) / (18 – 18)

= 0/0

Step IV: The function gives undefined form, apply L’hopital’s rule of limits.

lim_t→3 [(4t² – 9t – 9) / (2t² – 18)] = lim_t→3 [d/dt (4t² – 9t – 9) / d/dt (2t² – 18)]

= lim_t→3 [(8t^2-1 – 9(1) – 0) / (4t^2-1 – 0)]

= lim_t→3 [(8t – 9) / (4t)]

= lim_t→3 [(8t – 9)] /  [(4t)]

Step V: Substitute m=3 in the above expression again.

 [(4t² – 9t – 9) / (2t² – 18)] = (lim_t→3 [8t] – lim_t→3 [9]) / lim_t→3 [(4t)]

= ([8(3)] – [9]) / [4(3)]

= (24 – 9) / [12]

= 15 / 12

= 5/4

= 1.25

Use L’hopital’s rule calculator to solve the problems of this rule of limit calculus.

Summary

After reading the above post, you can solve any problem of limits by using its rules. Now you are witnessed that this topic is not difficult just a little effort is required. Once you grab all the basics of this post, you’ll master it.