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Home » News/Updates » Limit Calculus: A Brief Introduction with Examples
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Limit Calculus: A Brief Introduction with Examples

March 1, 2023No Comments5 Mins Read
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Limit Calculus A Brief Introduction with Examples

In 1647, the First time term “Limit” was used by Latin Mathematician Gregoire de Saint-Vincent (1584 – 1667).In 1908, English Mathematician G. H. Hardy (1877 – 1947) used the arrow symbol for limit in his book “A Course of Pure Mathematics”

Limit is an essential part of calculus. Limit plays an important role in integral, derivative, and continuity. In definite integral, we use limits as upper and lower limits. We can calculate the derivative by using the limit as Lim h→0 [f(x + h) – f(x)] / h. Continuity is also an application of limit.

In this article, we are going to discuss the definition of a limit, its properties, the Limit of the important functions, and some examples with step-by-step solutions.

Definition of limit

Let f(x) be a real-valued function, if “x” approaches a number “a” from the both left and right sides of “a” and f(x) approaches a specific number say “L” then “L” is called the limit of this function. The limit is written as 

Limx→a f(x) = L and read as “Limit of f(x), as “x” approaches “a” is “L”

Properties of limit

Some basic properties of the limit are given below: 

Let f and g be two function for which Lim x→a f(x) = A and Lim x→a f(x) = B, then

  • Addition property:

Lim x→a [f(x) + g(x)] = Lim x→a [f(x)] + Lim x→a [g(x)] = A + B

For example: Lim x→2 (x + 1) = Lim x→2 (x) + Lim x→2 (1) = 2 + 1 = 3

  • Subtraction property:

Lim x→ a [f(x) – g(x)] = Lim x→a [f(x)] – Lim x→a [g(x)] = A – B

For example: Lim x→1 (2x – 1) = Lim x→1 (2x) – Lim x→1 (1) = 2 – 1 = 1

  • Product property:

Lim x→a [f(x).g(x)] = Lim x→a [f(x)]. Lim x→a [g(x)] = A.B

For example: Lim x→2 (x – 1) (x + 1) = Lim x→2 (x – 1). Lim x→2 (x + 1) = (1) (3) = 3

  • Division property:

Lim x→a [f(x) / g(x)] = Lim x→a [f(x)] / Lim x→a [g(x)]

(Here limit of g(x) should be non-zero)

For example: Lim x→3 (5x) / (2) = Lim x→3 (5x) / Lim x→3 (2) = 15 / 3

  • Constant property:

Let c is any constant then 

Lim x→a [c f(x)] = c Lim x→a [f(x)] = c A

For example: Lim x→2 (3x) = 3 Lim x→2 (x) = (3) (2) = 6

  • Power property:

Lim x→a [f(x)] n = (A) n

Where n is any integer. 

See Also: 11th Class Maths Notes

Limit of some important function

Sometimes when we calculate the Limit we get the (0 / 0) form, in this condition simplify the function by using algebraic rules of making factors and then canceling the common factors if it is possible. Some important results to remember:

  • Lim x→ a [x n – an /x – a] = n an-1  (n belong to the set of integer and a greater than 0 )
  • Lim x→ +∞ (1 + 1 / n)n = e
  • Lim x→ 0  (ax – 1) / x = logea
  • Lim θ→ 0  (Sin θ / θ) = 1
  • Lim x→ ∞  (ex) = ∞
  • Lim x→ – ∞ (ex) = 0
  • Lim x→ ± ∞ (a / x) = 0 (where “a” is any real number)
  • Lim θ→ 0  (tan θ / θ) = 1
  • Lim θ → 0  Cos θ  = 1
  • Lim x → ∞ (1 + 1 / n )n = e

Solved Examples 

Example 1:

Evaluate the limit of the function 4x3 + 2x2 + 3x when x approaches 2.

Solution:

Step 1: Type the problem into the limit 

Lim x→2 (4x3 + 2x2 + 3x)

(If already written in the form of the limit then move to the next step)

Step 2: By using the addition property of limit, the above limit can be written as 

Lim x→2 (4x3) + Lim x→2 (2x2) + Lim x→2 (3x)

Step 3: By using the constant property of the limit, write the constant outside of the limit.

4 Lim x→2 (x3) + 2 Lim x→2 (x2) + 3 Lim x→2 (x)

Step 4: Now calculate the limit 

4 (23) + 2 (22) + 3 (2)

Step 5: Simplify 

32 + 8 + 6 = 46

A limits calculator can also be used as an alternate method of solving limit problems instead of performing lengthy calculations manually. 

Example 2:

Evaluate Lim x→2 (x2 – 4) / (x – 2)

Solution:

Lim x→2 (x2 – 22) / (x – 2)

We can observe if we take the limit we get the (0 / 0) form. Now use the algebraic technique. As we know that 

(a2 – b2) = (a – b) (a + b) put this formula in the above example so we get

Lim x→2 (x – 2) (x + 2) / (x – 2)

After further simplification we get 

Lim x→2 (x + 2) = 4

Example 3:

Lim θ→ 0 (tan 9θ / θ) 

Solution:

As we know that Lim θ→ 0 (tan 9θ / θ) = 1 

(Multiply and divide by 9)

Lim θ→ 0 (9tan 9θ / 9θ) 

9 Lim θ→ 0 (tan 9θ / 9θ) = 9(1) = 1

Conclusion 

In this article, we have discussed the definition of a limit, its applications, and the history of a limit in detail. Today we use arrow symbols in limit introduced by English Mathematician “G. H. Hardy”. Properties of limits with examples covered in this article. Further some limits of the important function are also discussed. 

In the example section, we have covered limit problems with step-by-step solutions for important functions. We learned how to solve limit problems when we get the (0 / 0) form. You observed that it is not a difficult topic after reading this article. You’ll able to deal with any problems related to the limit.

 

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Amir Shafiq

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