Introduction to Limit Calculus: A Brief Guide With Examples

 

Introduction to Limit Calculus: A Brief Guide With Examples

In calculus and mathematical analysis, limits play a very essential role in defining the Taylor series integrals and differentiation continuity. Limit is a well-known term for calculus and is very helpful in finding the function at a particular point. And also check the behavior of the function.

Limit is very useful in real life, like measuring the strength of the electric and gravitational fields.  In this section, we will discuss the definition of limit, examples, properties of limit, and also some special functions of limits

Definition of limit:

Let a function f (x) can be defined in an open interval near the number “a “, (need not be at a).

If, as x approaches a specific number “L” then l is called the limit of f(x) as x approaches a.

Symbolically it is a written Lim_x→a f(x) = L read as “limit of f(x), as x approaches a.

Types of limits:

Generally, there are three types of limits

•       Right-hand limit

•       Left-hand limit

•       Two side limits

Right-hand limit

A function f is said to have a right-hand limit if the Lim_x→c f(x) = M is read as the limit f(x) is equal to M as x approaches c from the right.

Formula

Left-hand limit

A function f is said to have a left-hand limit

Is read as the limit f(x) is equal to M as x approaches c from the left hand.

Formula

Two side limits

Two side limits appear when the left- and right-hand limit is equal.

Formula

=

The limit of the function is existing when the left- and the right-hand limit is existing otherwise not exist.

Properties of limits

First of all, suppose that two functions Lim_x→a f(x) = S and Lim_x→a g(x) = P exists and c is any constant. Below are a few properties of limit calculus that are helpful for determining limits problems.

Addition property

According to this property, the notation applied to each function separately. The equation for the additional property is:

Lim_x→a {f(x)+g(x)} = Lim_x→a f(x)+ Lim_x→a g(x) = S + P

For example

Lim_x→3 (2x+3x) = Lim_x→3 (2x) + Lim_x→3 (3x) (by using addition property)

Same as for subtraction.

Constant property

According to this property, the constant is written outside the limit because the limit is only applied to the variable. The equation for the constant property is:

Lim_x→a k f(x) = k Lim_x→a f(x) =KS

For example

Lim_x→3 (2x) =2 Lim_x→3(x) (by using constant property)

Product property

According to this property, a limit is applied to each function separately. The equation for the product property is: 

Lim_x→a f(x). g(x) = {Limx_→a f(x)}. {Lim_x→a g(x)} = S.P

For example

Lim_x→3(2x). (3x) = {Lim_x→3 (2x)}. {Lim_x→3 (3x)} (by using product property)

Quotient property

According to this property, a limit is applied to each function separately. The equation for the quotient rule i

Lim_x→a f(x)/g(x) = Lim_x→a f(x)/ Lim_x→a g(x) = S/P if P ≠ 0

For example

Lim_x→3 (2x)/(3x) = Lim_x→3 (2x)/ Lim_x→3(3x) (by using quotient property)

Power property

Lim_x→a {f(x)} ⁿ = {Limx_→a f(x)} ⁿ = Sⁿ

•       Where n is any integer

The particular function of limit

If, by substituting the number that x approaches into the function, we get (0/0), then we evaluate the limit as follows:

We simplify the given function by using the algebraic technique of making factors if possible and canceling the common elements. The method is explained in the following limits.

•       Lim_x→a (xⁿ – aⁿ) / (x – a) = na^n-1

_•            Lim_x→a   sinx/x = 1

•       Lim_x→0 e^x = 1

_•            Lim_x→+∞ (1 + (1/n)) ⁿ=e (when n tends to infinity through positive integral values only)

•       Lim _{x→ ∞} (e^x) = ∞

_•            Lim_x→-∞ (e^x)=0

_•            Lim_x→∞ (1 + (1/x)) ^x =e (as x approaches infinity while taking on or negative real values.)

_•            Lim_x→∞ (a/x) =0

Now we try to clear our topic by using some simple and easy examples of limits.

Solved Examples

Example: 1

Evaluate Lim_x→3 (9x² + 2x + 3) if it exists.

Solution:

Step 1:  first we apply the addition property on this function so,

Lim_x→3 (9x²) + Lim_x→3(2x) + Lim_x→3 (3)

Step 2:

Now use the constant property of the limit and write the constant outside of the limit.

9 Lim_x→3(x²) + 2 Lim_x→3 (x) + Lim_x→3(3)

Step 3:

Now apply the limit on the function

= 9 (3²) + 2(3) + 3

Step 4: by simplifying

81 + 6 + 3 = 90

Example: 2

Evaluate Lim_x→3 (x-3/√x -√3) if it exists.

Solution:

If we apply the direct limit on the function, the numerator, as well as the denominator, will become zero and we got the (0/0) form so, first, we factorize the function and then apply the limit

Step 1:

By making the factors of (x-3) 

Lim_x→3 (√x -√3) (√x +√3) / Lim_x→3 (√x -√3) (By using the quotient property)

Step 2:

Now the negative terms cancel each other and we get only the positive term.

Lim_x→3 (√x +√3)

Step 3:

Now we apply additional property so,

Lim_x→3 (√x) + Lim_x→3(√3)

Step 4:

Now apply the limit we get,

 = (√3) + (√3)

= 2 √3

Example 3:

Evaluate Lim_x→∞ (1+1/9x)^9x if it exists.

Solution:

We know that Limx→∞ (1+1/x)^x = e by the particular function of limit.

Lim_x→∞ (1+1/9x)^9x = e

Conclusion:

 As we have seen limit is a fundamental concept in mathematics and also for calculus. In this article, we try to clear the definition, types, and some properties of limit. And try to understand the topic through some examples. And also learn how to solve the intermediate form of function.