Limits are frequently used in mathematics for solving the numerical value of the function at a particular point. It is a well-known branch of calculus that is very essential for evaluating the differential, continuity, and integral of the function.
Limit plays an important role in solving the differential by the first principle method, finding the area by applying the upper and lower limit in definite integral, to check whether the function is continuous or not.
Limits Definition
The limit of the function is defined as the “limit of f(x), as “x” approaches to “a” from both sides (left and right-hand side), equals to M.”
Lim_{x}_{→a} f(x) = M
Or
Lim_{x}_{→a}^{+} f(x) = Lim_{x}_{→a}^{– }f(x) = M
This means that the limit of the function exists when both the left-hand side and the right-hand side of the limit become equal. When the specific point comes from the left or right then the left-hand limit and right-hand limit are obtained.
A limit calculator by Allmath is an online resourece to evaluate the limits of functions according to the definition.
The limit of the functions can be evaluated with the help of the laws of limit. Now we are going to explain the laws of limit calculus.
Laws of limit calculus
Below are a few well-known laws of limit calculus.
Laws Name |
Definition |
Expression |
Law of sum |
The law of sum is used when there are two or more terms or functions with a sum notation between them. The expression of the law of sum is written as: |
Lim_{y}_{→}_{a} [h(y) + g(y)] = Lim_{ys}_{→}_{a} [h(y)] + Lim_{y}_{→}_{a} [g(y)] |
Law of constant |
Law of constant states that the constant function remains the same after applying the limit values as there is no independent variable present in the function. The expression of the law of constant is written as: |
Lim_{y}_{→}_{a} [C] = C |
Law of constant of function |
According to the law of constant of function, the constant coefficients along with the functions will come outside the limit notation. |
Lim_{y}_{→}_{a} [C h(y)] = C Lim_{y}_{→}_{a} [h(y)] |
Law of difference |
The law of difference is used when there are two or more terms or functions with a minus notation between them. |
Lim_{y}_{→}_{a} [h(y) – g(y)] = Lim_{y}_{→}_{a} [h(y)] – Lim_{y}_{→}_{a} [g(y)] |
Law of product |
The law of product is used when there are two or more terms or functions with a product notation between them. The expression of the law of product is written as: |
Lim_{y}_{→}_{a} [h(y) * g(y)] = Lim_{y}_{→}_{a} [h(y)] * Lim_{y}_{→}_{a} [g(y)] |
Law of division |
The law of division is used when there are two or more terms or functions with a quotient notation between them. The expression of the law of division is written as: |
Lim_{y}_{→}_{a} [h(y) / g(y)] = Lim_{y}_{→}_{a} [h(y)] / Lim_{y}_{→}_{a} [g(y)] |
Law of Power |
According to this law, the power of the function will be applied after applying the specific point. |
Lim_{y}_{→}_{a} [h(y)]^{n} = [Lim_{y}_{→}_{a} h(y)]^{n} |
Law of L’hopital’s |
To get rid of the undefined form of the function, this law is used. The expression of this law is: |
Lim_{y}_{→}_{a} [h(y) / g(y)] = Lim_{y}_{→}_{a} [d/dy (h(y)) + d/dy (g(y))] |
Laws of trigonometry |
The well-known trigonometry laws are: |
1. Lim_{y}_{→}_{0} [sin(y) / y] = 1 2. Lim_{y}_{→}_{0} [tan(y) / y] = 1 3. Lim_{y}_{→}_{0} [1 – cos(y) / y] = 0 4. Lim_{y}_{→}_{0} [sin(y)] = 0 5. Lim_{y}_{→}_{0} [e^{y}] = 1 6. Lim_{y}_{→}_{0} [cos(y)] = 1 |
Methods of evaluating limits
There are various methods of solving limits such as:
1. Direct substitution method: Direct put the value of a specific point to the function at the place of the independent variable with the help of laws.
2. Factoring and canceling method: Make the factors of the given function and cancel the similar functions to the numerator and denominator.
3. Fraction within fraction methods
4. Conjugate method: when the square root function is given, take the conjugate of the function.
Expansion method
These methods are very essential for evaluating the problems of limit calculus to find the approaching value of the limit.
How to evaluate limits?
The following examples will be helpful to learn how to evaluate limits.
Example 1: For the direct substitution method
Find the limit at “2” of the functionf(w) = (2w^{3}+ sin(w) + 18) / (6w^{4} – 3w^{2} + 2).
Solution
Step 1:First of all, write the given function according to definition of limit.
Lim_{w}_{→}_{a }[f(w)] = lim_{w}_{→2} [(2w^{3} + sin(w) + 18) / (6w^{4} – 3w^{2} + 2)]
Step 2:
Apply the laws of sum, difference, and division of limit calculus to the above expression.
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = (lim_{w}_{→2 }[2w^{3}] + lim_{w}_{→2 }[sin(w)] + lim_{w}_{→2 }[18]) / (lim_{w}_{→2 }[6w^{4}] – lim_{w}_{→2 }[3w^{2}] + lim_{w}_{→2 }[2])]
According to the law of constant of function.
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = (2lim_{w}_{→2 }[w^{3}] + lim_{w}_{→2 }[sin(w)] + lim_{w}_{→2 }[18]) / (6lim_{w}_{→2 }[w^{4}] – 3lim_{w}_{→2 }[w^{2}] + lim_{w}_{→2 }[2])]
Step 3:
Now apply “2” to the above expression.
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = (2 [2^{3}] + [sin(2)] + [18]) / (6 [2^{4}] – 3 [2^{2}] + [2])
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4 }– 3w^{2} + 2)] = (2 [8] + [sin(2)] + [18]) / (6 [16] – 3 [4] + [2])
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = (16 + [sin(2)] +18) / (96 – 12 +2)
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = (34 +sin(2)) / (84 +2)
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = (34 +sin(2)) / (82)
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = 34/82 +sin(2) /82
lim_{w}_{→2 }[(2w^{3} + sin(w)/w + 18) / (6w^{4} – 3w^{2} + 2)] = 17/41 +sin(2) /82
Example 2: For factoring and canceling
Find the limit at “1” of the function f(y) = (2y^{2 }– 2y – 1)/(4y^{2} – 1).
Solution
Step 1: First of all, write the given function according to definition of limit.
lim_{y}_{→}_{a }[f(y)] = lim_{y}_{→1} [((2y^{2} – y – 1)/(4y^{2 }– 1)]
Step 2: Now make the factors of the given expression.
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = lim_{y}_{→1 }[((2y^{2} – 2y + y – 1)/((2y)^{2} – (1)^{2})]
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = lim_{y}_{→1 }[((2y (y – 1) + 1 (y – 1))/((2y – 1) (2y + 1))]
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = lim_{y}_{→1 }[((2y + 1) (y – 1))/((2y – 1) (2y + 1))]
Step 3: Now cancel the similar terms from the above expression.
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = lim_{y}_{→1 }[((2y + 1) (y – 1))/((2y – 1) (2y + 1))]
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = lim_{y}_{→1 }[((y – 1))/((2y – 1))]
Step 4: Now apply the limit.
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = ((lim_{y}_{→1 }[y] – lim_{y}_{→1} [1]))/((lim_{y}_{→1 }[2y] – lim_{y}_{→1 }[1])
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = (([1] – [1]))/([2 (1)] – [1])
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = ((1 – 1))/([2 –1])
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = (0)/(1])
lim_{y}_{→1 }[((2y^{2} – y – 1)/(4y^{2} – 1)] = 0
Conclusion
Limit is a main topic of calculus that is frequently used to evaluate the approaching value of the given functions at the specific point. The laws of limit plays an essential role in solving the limit problems and there are various method to evaluate the limits of a function.