Limit problems: How to Calculate Them by Using Laws of Limit Calculus?

In calculus, limit is one of its types widely used to define the other branches like differential, integral, Taylor series, Maclaurin series, continuity, etc. Limits play a vital role in calculus for various purposes. 

It is used to find the numerical value of the functions at specific points. The specific point of the limit calculus could be finite or infinite. The problems of limit calculus are to be evaluated with the help of laws of limits. 

What is the limit in calculus?

Limits of functions are defined as “limits of f(x) when x approaches “a” from the left and the right side of the curve, equals N.”

The limit is a well-known branch of calculus used to evaluate the numerical value of the function at a particular point. Basically, it is used to define continuity, derivative, and integral calculus.

In derivatives, limits are used to find the differential of the functions with respect to an independent variable using the limit through the first principle method. In integrals, the limits are used as boundary values for the definite integral to evaluate the area under the curve. 

In continuity, the limits are used to check the continuity and discontinuity of the functions. Hence, limits are very essential in calculus as it is very helpful for evaluating the various kinds of problems and dealing with the other branches of calculus.

The limits are of three types such as left-hand limit, right-hand limit, and two-sided limits. 

Laws of limit calculus

The laws of limit calculus are essential for dealing with the problems of limit. 

The law of sum is used when the function with two or more terms are given with the plus sign. According to this law, the notation of limit calculus is applied with each term of the function separately with the sign of addiction among them. 

Lim_v→a [h(v) + g(v)] = Lim_v→a [h(v)] + Lim_v→a [g(v)]

The law of difference is used when the function with two or more terms are given with the minus sign. According to this law, the notation of limit calculus is applied with each term of the function separately with the sign of difference among them. 

Lim_v→a [h(v) - g(v)] = Lim_v→a [h(v)] - Lim_v→a [g(v)]

The law of difference is used when the function with one or more terms are given with the power. According to this law, the exponent is applied after applying the specific point of limit. 

Lim_v→a [h(v)]ⁿ = [Lim_v→a [h(v)]]ⁿ 

Limits are not applied to the constant function as there is no independent variable present in the given function. 

Lim_v→a [C] = C

The law of constant function is applied when the constant coefficients are present along with the function. According to this law, the constant term will be written before the limit notation. 

Lim_v→a [C * h(v)] = C * Lim_v→a [h(v)]

The law of product is used when the function with two or more terms are given with the multiply sign. According to this law, the notation of limit calculus is applied with each term of the function separately with the sign of multiply among them. 

Lim_v→a [h(v) * g(v)] = Lim_v→a [h(v)] * Lim_v→a [g(v)]

The law of division is used when the function with two or more terms are given with the division sign. According to this law, the notation of limit calculus is applied with each term of the function separately with the sign of division among them. 

Lim_v→a [h(v) / g(v)] = Lim_v→a [h(v)] / Lim_v→a [g(v)]

This law is used when the function is undefined. According to this law, you have to take the differential of the numerator and denominator with respect to the independent variable. 

Lim_v→a [h(v) / g(v)] = Lim_v→a [d/dv [h(v)] / d/dv [g(v)]

A limit calculator by Allmath is a helpful way to evaluate the limit problems with steps according to the above laws of limit calculus. 

How to evaluate the limit of different kinds of calculus functions?

The following examples will be helpful to learn how to evaluate limits. 

Example 1: 

Find the limit value at the specific point “3” if the calculus function is:

h(v) = 4v⁴ – 2v + 18v² / 6v⁴ + 2v² + 2sin(v)

Solution 

Step 1: First of all, write the given expression of calculus according to the general expression of limit.  

Lim_v→a [f(v)] = lim_v→3 [4v⁴ – 2v + 18v² / 6v⁴ + 2v² + 2sin(v)]

Step 2: Apply the laws of sum, difference, and division to the terms of the above expression.

lim_v→3 [4v⁴ – 2v + 18v² / 6v⁴ + 2v² + 2sin(v)] = lim_v→3 [4v⁴] – lim_v→3 [2v] + lim_v→3 [18v²] / lim_v→3 [6v⁴] + lim_v→3 [2v²] + lim_v→3 [2sin(v)]

Step 3: Use the law of constant function to take out the constant coefficients outside the limit notation. 

lim_v→3 [4v⁴ – 2v + 18v² / 6v⁴ + 2v² + 2sin(v)] = 4lim_v→3 [v⁴] – 2lim_v→3 [v] + 18lim_v→3 [v²] / 6lim_v→3 [v⁴] + 2lim_v→3 [v²] + 2lim_v→3 [sin(v)]

Step 4: Now place the specific point of limit calculus to the above expression. 

lim_v→3 [4v⁴ – 2v + 18v² / 6v⁴ + 2v² + 2sin(v)] = 4 [3⁴] – 2 [3] + 18 [3²] / 6 [3⁴] + 2 [3²] + 2 [sin(3)]

= 4 [81] – 2 [3] + 18 [9] / 6 [81] + 2 [9] + 2 [sin(3)]

= 324 – 6 + 162 / 486 + 18 + 2 [sin(3)]

= 324 – 6 + 0.34 + 18 + 2 [sin(3)]

= 318 + 0.34 + 18 + 2 [sin(3)]

= 318.34 + 18 + 2 [sin(3)]

= 336.34 + 2 [sin(3)]

= 336.34 + 2 [0.14]

= 336.34 + 0.28

= 336.62

Example 2: 

Find the limit at “3” of the function, f(u) = (2u² – 3u – 9)/(4u² – 4u – 24). 

Solution 

Step 1: First of all, write the given expression of calculus according to the general expression of limit.  

Lim_u→a [f(u)] = lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24)]

Step 2: Apply the laws of sum, difference, and division to the terms of the above expression.

lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24)] = (lim_u→3 [2u²] – lim_u→3 [3u] – lim_u→3 [9])/( lim_u→3 [4u²] – lim_u→3 [4u] – lim_u→3 [24])

Step 3: Use the law of constant function to take out the constant coefficients outside the limit notation. 

lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24)] = (2lim_u→3 [u²] – 3lim_u→3 [u] – lim_u→3 [9])/(4lim_u→3 [u²] – 4lim_u→3 [u] – lim_u→3 [24])

Step 4: Now place the specific point of limit calculus to the above expression. 

lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24)] =  (2 [3²] – 3 [3] – [9])/(4 [3²] – 4 [3] – [24])

= (2 [9] – 3 [3] – [9])/(4 [9] – 4 [3] – [24])

= (18 – 9 – 9)/(36 – 12 – 24)

= (18 – 18)/(36 – 36)

= 0/0

Step 5: Use the law of L’hopital’s as the given function makes an undefined form.

lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24 = lim_u→3 [d/du (2u² – 3u – 9)/d/du (4u² – 4u – 24)]

lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24 = lim_u→3 [(4u – 3 – 0)/ (8u – 4 – 0)]

lim_u→3 [(2u² – 3u – 9)/(4u² – 4u – 24 = lim_u→3 [(4u – 3)/ (8u – 4)]

Step 6: Apply the specific value again.

= (lim_u→3 [4u] – lim_u→3 [3])/ (lim_u→3 [8u] – lim_u→3 [4])

= ([4(3)] – [3])/ ([8(3)] – [4])

= (12 – 3)/ (24 – 4)

= 9/20

Sum Up

The limit is a well-known branch of calculus used to evaluate the values of the functions at specific points. The laws of limit calculus are very essential for the calculations of the limit problems.