Limit is a well-known branch of calculus that is helpful for evaluating the numerical value of a function at a specific point and for defining the derivative calculus, integral calculus, continuity, and Taylor series.
Limit calculus is mostly helpful for finding the differential of a function through the first principle method, for finding the integral of the function the upper and lower limits are applied to the given function, and it is the main step to check the continuity of the function.
In calculus, the value that a function approaches as the function goes closer and closer to a specific point as input is said to be the limit of a function. The term limit calculus is a great idea that covers all the basics of calculus.
The function f(u) is said to be a limit of a function if the given function approaches a particular point that gives a numerical value is said to be a limit of a function. Mathematically it can be written as:
Limu→b f(u) = M
There are three main branches of limit calculus such as left-hand limit (the specific point if the limit comes from the left side of the curve), right-hand limit (the specific point if the limit comes from the right side of the curve), and two-sided limit (when both left-hand limit and right-hand limit becomes equal this type of limit exists).
The rules of limit calculus are very essential for finding the numerical value of the functions when they contain two or more terms. The function could be linear, algebraic, polynomial, trigonometric, logarithmic, exponential, or constant.
Rules Name |
General expression |
Theory |
Rule of Addition |
Limv→b [f(v) + h(v)] = Limv→b [f(v)] + Limv→b [h(v)] |
The rule of addition is used when two or more terms in a function are present with a positive sign between them. According to this rule, the notation of limit will be applied to each function separately. |
Rule of subtraction |
Limv→b [f(v) - h(v)] = Limv→b [f(v)] - Limv→b [h(v)] |
The rule of subtraction is used when two or more terms in a function are present with a negative sign between them. According to this rule, the notation of limit will be applied separately to each term of function. |
Rule of exponent |
Limv→b [f(v)]m = [Limv→b [f(v)]]m |
The rule of exponent also known as the power rule is used when the function is given along with an exponent term. |
Rule of multiplication |
Limv→b [f(v) x h(v)] = Limv→b [f(v)] x Limv→b [h(v)] |
The rule of multiplication is used when two or more terms in a function are present with a multiply sign (x, *) between them. According to this rule, the notation of limit will be applied separately to each term of function. |
Rule of division |
Limv→b [f(v) / h(v)] = Limv→b [f(v)] / Limv→b [h(v)] |
The rule of division is used when two or more terms in a function are present with a division sign (/,➗) between them. According to this rule, the notation of limit will be applied separately to each term of function. |
Rule of a constant function |
Limv→b [K * f(v)] = K * Limv→b [f(vs)] |
When a function has two or more terms along with a constant coefficient, then the law of constant function is used. |
L’hopital’s Rule |
Limv→b [f(v) / h(v)] = Limv→b [d/dv f(v)/ d/dv h(v)] |
The rule of L’hopital’s is applied when a function gives an undefined form. According to this rule, the function will be differentiated with respect to the independent variable. |
The limit of the function should be evaluated with the help of the following steps or methods.
You can take assitance from online tools like limit calculator by Allmath to evaluate the limits according to the above steps
Here are a few examples to learn how to find the limit of the function.
Example 1:
Apply the particular point “4” to the given function to find the numerical value of the function.
f(v) = 2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v
Solution
Step 1:First of all, identify the independent variable of the function and use the notation of limit to the given function.
Independent variable = v
Limv→b [f(v)] = limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v]
Step 2: Now apply the rules of addition, subtraction, and division to the given function.
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = limv→4 [2v2] – limv→4 [11v3] + limv→4 [4v] / limv→4 [2v5] – limv→4 [4v2] + limv→4 [12v]
Step 3: Now take out the constant coefficients outside the limit notation.
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = 2limv→4 [v2] – 11limv→4 [v3] + 4limv→4 [v] / 2limv→4 [v5] – 4limv→4 [v2] + 12limv→4 [v]
Step 4: Now substitute 4 in the place of “v”.
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = 2 [42] – 11 [43] + 4 [4] / 2 [45] – 4 [42] + 12 [4]
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = 2 [16] – 11 [64] + 4 [4] / 2 [1024] – 4 [16] + 12 [4]
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = 32 – 704 + 16 / 2048 – 64 + 48
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = 32 – 704 + 0.0078 – 64 + 48
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = -672 + 0.0078 – 64 + 48
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = -672.0078 – 64 + 48
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = -736.0078 + 48
limv→4 [2v2 – 11v3 + 4v / 2v5 – 4v2 + 12v] = – 688.0078
Example 2:
Apply the specific point “1” to the given function to determine the numerical value of the function.
f(u) = (6u2 – 2u – 4)/(12u2 + u – 13)
Solution
Step 1:First of all, identify the independent variable of the function and use the notation of limit to the given function.
Independent variable = u
Limu→b [f(u)] = limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)]
Step 2: Now apply the rules of addition, subtraction, and division to the given function.
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (limu→1 [6u2] – limu→1 [2u] – limu→1 [4])/( limu→1 [12u2] + limu→1 [u] – limu→1 [13])
Step 3: Now take out the constant coefficients outside the limit notation.
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (6limu→1 [u2] – 2limu→1 [u] – limu→1 [4])/(12limu→1 [u2] + limu→1 [u] – limu→1 [13])
Step 4: Now substitute 1 in the place of “u”.
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (6 [12] – 2 [1] – [4])/(12 [12] + [1] – [13])
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (6 [1] – 2 [1] – [4])/(12 [1] + [1] – [13])
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (6 – 2 – 4)/(12 + 1 – 13)
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (4 – 4)/(13 – 13)
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (0)/(0)
Step 5:Use the law of L’hopital’s as the given function makes undefined form.
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = limu→1 [d/du (6u2 – 2u – 4)/ d/du (12u2 + u – 13)]
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = limu→1 [(12u – 2(1) – 0)/ (24u + 1 – 0)]
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = limu→1 [(12u – 2)/ (24u + 1)]
Step 6: Apply the specific value again.
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (limu→1 [12u] – limu→1 [2])/ (limu→1 [24u] + limu→1 [1])
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (12(1) – [2])/ (24(1) + [1])
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (12 – 2)/ (24 + 1)
limu→1 [(6u2 – 2u – 4)/(12u2 + u – 13)] = (10)/ (25) = 2/5
Limit calculus is very essential in calculus for defining other branches like differential, integral, and continuity. It is helpful for finding the value of the function at a specific point. The problems of the limit are to be evaluated with the help of rules of limit calculus.