Limit in Calculus - Explained with Its Definition Types and Calculations
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Limit in Calculus - Explained with Its Definition Types and Calculations
Limit in Calculus - Explained with Its Definition Types and Calculations
Limit in Calculus - Explained with Its Definition Types and Calculations
Updated On:2022-11-18 09:44:33
In mathematics, the limit is a value that a function (or sequence) "approaches" as the input (or index) "approaches" some value. Limits are important in calculus and analysis in general.
Most of the limits that occur in practice have already been calculated and can be looked up in mathematical tables. In this blog post, we will go over the definition of a limit, the types of limits, how to calculate limits, and finally some examples of calculating limits.
The Definition of a Limit
The definition of a limit is a numerical value that a function approaches as the input gets closer and closer to some value. The limit of the functions is represented in a general expression to evaluate its numerical result. The general expression is:
- “w” is the independent variable of the function.
- “f(w)” is the function whose limit value is to be evaluated.
- “a” is the particular point that a function approaches.
- “L” is the numerical result of the function after applying the limit value.
Types of limits in calculus
There are three types of limits:
- One-sided limit
- Two-sided limit
- Infinite limit
Let us discuss the types of limits in calculus briefly.
One-sided Limits
A one-sided limit is defined as the maximum value that a function can reach when the input is close to but not exactly at that point.
For example, if you have a function that takes an input and returns an output, then its one-sided limit would be infinity (i.e., it can never reach this maximum value).
Two-Sided Limits
A two-sided limit is defined as the minimum and maximum values that a function can reach when the input is close to but not exactly at that point.
For example, if you have a function that takes an input and returns an output, then its two-sided limit would be -1 (i.e., it could theoretically reach this minimum or maximum value). Similarly, if you have a function that takes a float input and returns a floating output, then its two-sided limit would be 0 (i.e., it could theoretically reach this minimum or maximum value).
Infinite Limits
An infinite limit is simply defined as any number beyond which there's no known way to calculate the corresponding limit using standard methods like algebraic manipulation or the Sandwich Theorem.
For example, consider pi – 3 degrees Celsius: there's no known finite amount above which pi will increase by 3 degrees Celsius; instead, it continues on forever towards infinity. So technically speaking, pi has an infinite limit (although we don't know what it actually equals!).
How to calculate limits in calculus?
To calculate a limit, you can use different methods depending on the type of limit you're dealing with. For one-sided limits, you can use substitution or factoring; for two-sided limits, you can use algebraic manipulation or graphing;
and for infinite Limits, you can use L’hopital’s rule.A L’hopital’s rule is a technique for calculating the numerical values of the indeterminate functions by taking the differential of the numerator and denominator and applying the value of the limit.
If the function again formsan indeterminate form, you have to take the second derivative of the function, and so on.
Let us take a few examples of limit calculus solved with the help of the laws of limit calculus.
Example 1:
Evaluate the limit of the given function if the particular point “w” approaches 3.
f(w) = 6w + 12w4 – 2w3 + 12
Solution
Step 1:First of all, apply the notation of limit calculus to the given function along with the value of the specific point.
Limw→af(w) = Limw→3[6w + 12w4 – 2w3 + 12]
Step 2:Now use the sum and difference laws of limit calculus to apply the notation of limit with each term of the function separately and take out the constant coefficients.
Limw→3[6w + 12w4 – 2w3 + 12] = Limw→3[6w] + Limw→3[12w4] – Limw→3[2w3] + Limw→3[12]
Limw→3[6w + 12w4 – 2w3 + 12] = 6Limw→3[w] + 12Limw→3[w4] – 2Limw→3[w3] + Limw→3[12]
Step 3:Now put the value of the particular point in the place of the independent variable of the above expression.
Limw→3[6w + 12w4 – 2w3 + 12] = 6[3] + 12[34] – 2[33] +[12]
= 6[3] + 12[3 x 3 x 3 x 3] – 2[3 x 3 x 3] +[12]
= 6[3] + 12[81] – 2[27] +[12]
= 18 + 972 – 54 +12
= 990 – 54 +12
= 936 +12
= 948
A limit calculator with steps can be used to evaluate the limits to ease up the lengthy and time consuming calculations
Example 2:
Calculate the limit of the given function if the particular point “w” approaches 2.
h(t) = (2t3 – 2t2 – 8) / (t2 – 22)
Solution
Step 1:First of all, apply the notation of limit calculus to the given function along with the value of the specific point.
Limt→ah(t) = Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)]
Step 2:Now use the sum and difference laws of limit calculus to apply the notation of limit with each term of the function separately and take out the constant coefficients.
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (Limt→2[2t3] – Limt→2[2t2]– Limt→2[8]) / (Limt→2[t2] – Limt→2[22])
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (2Limt→2[t3] – 2Limt→2[t2]– Limt→2[8]) / (Limt→2[t2] – Limt→2[22])
Step 3:Now put the value of the particular point in the place of the independent variable of the above expression.
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (2[23] – 2[22]–[8]) / ([22] –[22])
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (2[8] – 2[4]–[8]) / ([4] –[4])
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (16 – 8–8) / (4 –4)
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = 0/0
Step 4:Now use L’hopital’s rule of limit calculus as the given function makes an indeterminate form.
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = Limt→2[d/dt (2t3 – 2t2 – 8) / d/dt (t2 – 22)]
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = Limt→2[(6t2 – 4t– 0) / (2t – 0)]
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = Limt→2[(6t2 – 4t) / (2t)]
Step 5: Now again put the value of the particular point in the place of the independent variable of the above expression.
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (Limt→2[6t2] – Limt→2[4t]) / (Limt→2[2t])
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (6(2)2 – 4(2)) / (2(2))
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (6(4) – 4(2)) / (2(2))
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (24 – 8) / (4)
Limt→2[(2t3 – 2t2 – 8) / (t2 – 22)] = (16) / (4) = 4
Example 3:
Find the limit of the given function if the particular point “w” approaches 1.
f(u) = 2u2 / 3u * 12u3 / 2u
Solution
Step 1:First of all, take the given function p(r) and apply the limit notation to it.
Limu→af(u) = Limu→1[2u2 / 3u * 12u3 / 2u]
Step 2:Now apply the limit notation to each function separately with the help of product and quotient laws of limit calculus.
Limu→1[2u2 / 3u * 12u3 / 2u] = Limu→1[2u2] / Limu→1[3u] * Limu→1[12u3] / Limu→1[2u]
Limu→1[2u2 / 3u * 12u3 / 2u] = 2Limu→1[u2] / 3Limu→1[u] * 12Limu→1[u3] / 2Limu→1[u]
Step 3:Now substitute r = 1 to the above expression.
Limu→1[2u2 / 3u * 12u3 / 2u] = 2[12] / 3[1] * 12[13] / 2[1]
= 2[1] / 3[1] * 12[1] / 2[1]
= 2 / 3 * 12 / 2
= 0.67 * 12 / 2
= 8 / 2
= 4
Wrap up
Now you can get all the basics of limits in calculus from this post. We have discussed all the basics of limits such as their definition, types, and calculation along with examples. Once you go through this post, you are able to solve any problem of limit.
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