What is the method for evaluating the Limits of Functions?
Explained with examples
In calculus, the limit is a wide concept that is frequently used to calculate the numerical values
of the functions and for defining the other branches of calculus to ease up the calculation. It
can define the Taylor series, derivative, integration, and other branches of calculus.
In differentiation, the limits are used in the first principle method to differentiate the function
by applying limits. In this lesson, we will learn the concept of how to evaluate the limit
problems by defining the limits, rules, and solved problems.
What is the limit in calculus?
In calculus, the limit is a term that offers (allows) the given function to approach some numeric
values as the given expression of that function that goes closer and closer to the specific value.
The corresponding variable is essential in limits as the value of the specific point is to be a
substitute for the corresponding variable available in the function.
Limit’s Formula in Calculus
In calculus, the formula or general expression for denoting the limit is:
Limt→s f(t) = N
This expression of limit calculus can be read as the limit of the function f(t) as t approaches s is
equal to N.
In the formula of limit,




t is the corresponding variable of the function
s is the specific point of the function.
f(t) is the given function
N is the numerical result of the function
Laws of limit in calculus
Here are some basic rules (laws) of limits in calculus that are used to solve the problems of
limits easily.
Name of the law
Sum Law
Difference Law
Constant Law
Constant function Law
Power Law
Product Law
Quotient rule
L’hopital’s rule
Law
Limt→s [f(t) + g(t)] = Limt→s [f(t)] + Limt→s [g(t)]
Limt→s [f(t) - g(t)] = Limt→s [f(t)] - Limt→s [g(t)]
Limt→s [k] = k
Limt→s [k * f(t)] = k * Limt→s [f(t)]
Limt→s [f(t)]2 = [Limt→s [f(t)]2
Limt→s [f(t) * g(t)] = Limt→s [f(t)] * Limt→s [g(t)]
Limt→s [f(t) / g(t)] = Limt→s [f(t)] / Limt→s [g(t)]
Limt→s [f(t) / g(t)] = Limt→s [f’(t) / g’(t)]
How to evaluate the limit of the functions?
Let us take a few examples of evaluating the limits of the functions by using its rules.
Example 1: For L’hopital’s rule
Evaluate the limit of f(w) = w3 – 1 / (4w2 + 2w – 6) by using the L’hopital’s rule if the particular
value is 1.
Solution
Step 1: First of all, apply the notation of limit to the given function.
Limw→s f(w) = Limw→1 [w3 – 1 / (4w2 + 2w – 6)]
Step 2: Now write the notation of limit with each term of the function separately with the help
of the sum and difference rule.
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = [Limw→1 w3 – Limw→1 1 / (Limw→1 4w2 + Limw→1 2w – Limw→1
6)]
Step 3: Now put the value of the specific point to the above function.
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = [13 – 1 / (4(1)2 + 2(1) – 6)]
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = [1 – 1 / (4(1) + 2(1) – 6)]
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = [1 – 1 / (6 – 6)] = 0/0
Step 4: Apply L’hopital’s rule of limit as the function gives 0/0 form. Now find the derivative of
the function and apply the limit value again.
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = Limw→1 [d/dw (w3 – 1) / d/dw (4w2 + 2t – 6)]
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = Limw→1 (3w2 – 0) / (8w + 2 – 0)]
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = Limw→1 (3w2) / (8w + 2)]
Now substitute w = 1
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = (3(1)2) / (8(1) + 2)]
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = [3 / 8 + 2]
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = 3/10
Limw→1 [w3 – 1 / (4w2 + 2w – 6)] = 0.3
Example 2: For sum and difference rule
Evaluate the limit of f(z) = 12z4 – 16z3 + 3z2 + 15 if the particular value is 2.
Solution
Step 1: First of all, apply the notation of limit to the given function.
Limz→s f(z) = Limz→2 [12z4 – 16z3 + 3z2 + 15]
Step 2: Now write the notation of limit with each term of the function separately with the help
of the sum and difference rule and take out the constant coefficients outside the limit notation.
Limz→2 [12z4 – 16z3 + 3z2 + 15] = Limz→2 [12z4] – Limz→2 [16z3] + Limz→2 [3z2] + Limz→2 [15]
Limz→2 [12z4 – 16z3 + 3z2 + 15] = 12Limz→2 [z4] – 16Limz→2 [z3] + 3Limz→2 [z2] + Limz→2 [15]
Step 3: Now put the value of the specific point to the above function.
= 12 [24] – 16 [23] + 3 [22] + [15]
= 12 [2 x 2 x 2 x 2] – 16 [2 x 2 x 2] + 3 [2 x 2] + [15]
= 12 [16] – 16 [8] + 3 [4] + [15]
= 192 – 128 + 12 + 15
= 64 + 12 + 15
= 76 + 15
= 91
A limit calculator can be used to calculate the above problem to avoid time-consuming
calculations.
Example 3: For product and quotient rule
Evaluate the limit of p(t) = 12t3 * 6t2 / 21t4 * 15t3 * 5 by using the product and quotient laws if
the particular value is 3.
Solution
Step 1: First of all, take the given function and apply the limit notation and particular value to it.
p(t) = 12t3 * 6t2 / 21t4 * 15t3 * 5
corresponding value = t = 3
Limt→s [p(t)] = Limt→3 [12t3 * 6t2 / 21t4 * 15t3 * 5]
Step 2: Now write the notation of limit with each term of the function separately with the help
of quotient and product rule and take out the constant coefficients outside the limit notation.
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = Limt→3 [12t3] * Limt→3 [6t2] / Limt→3 [21t4] * Limt→3 [15t3] *
Limt→3 [5]
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 12Limt→3 [t3] * 6Limt→3 [t2] / 21Limt→3 [t4] * 15Limt→3 [t3] *
Limt→3 [5]
Step 4: Now substitute the value of the corresponding variable w = 2 to the above expression in
order to calculate the limit result of the given function.
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 12 [33] * 6 [32] / 21 [34] * 15 [33] * [5]
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 12 [3 x 3 x 3] * 6 [3 x 3] / 21 [3 x 3 x 3 x 3] * 15 [3 x 3 x 3] *
[5]
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 12 [27] * 6 [9] / 21 [81] * 15 [27] * [5]
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 324 * 54 / 1701 * 405 * 5
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 17496 / 1701 * 405 * 5
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 10.29 * 405 * 5
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 4167.45 * 5
Limt→3 [12t3 * 62 / 21t4 * 15t3 * 5] = 20837.25
Final Words
Now you are witnessing that calculating the limit of the functions is not a difficult task. You can
grab all the basics of calculating the limit of the functions from this lesson by learning the rules
and solved examples.