Definition of Limit Calculus

A limit is a mathematical computation that tells us the value assumed by a mathematical expression/function as the independent variable approaches a certain value.


Limit of a Function


Let a function `f(x`) be defined in an open interval in the neighborhood of the number "a".
If as `x` approaches "a" from both left and right sides of "a", `f(x`)approaches a specific number "L" then "L" is called the limit of `f(x)` as `x` approaches a.


Symbolically,
Limit is expressed as follows:

`Lim_(x->a) f(x) = L `


And is read as "Limit of `f(x)` as `x-> `a is L" .


Under this definition of the limit of a function, we can categorize limits into these three types:


  •  Limit of a function as X approaches some fixed constant
  •  Limit of a function as X approaches +ve/-ve infinity
  •  Limit of a function as X approaches 0

There are some fundamental theorems on Limits of functions. We present their brief statements here.


Fundamental Theorems on Limits of functions


1. Let us have two functions `u` and `v` for which

` Lim_(x->a) u(x)= P` and `Lim_(x->a) v(x) = Q` then
`Lim_(x->a) [u(x) + v(x)] `= `Lim_(x->a) u(x) `+ `Lim_(x->a) v(x) `= P + Q

Example 1:

Explanation:

`Lim_(x->4) (x + 6)` = `Lim_(x->4) x` + `Lim_(x->4) 6` = 4 + 6 = 10

2. For the two functions u and v discussed above,

`Lim_(x->a) [u(x) – v(x)] `= `Lim_(x->a) u(x)` – `Lim_(x->a) v(x) `= P – Q

Example 2:

Explanation:

`Lim_(x->4) (x – 6)` = `Lim_(x->4) X` – `Lim_(x->4) 6` = 4 – 6 = –2

3. Let us have a real number c,then

`Lim_(x->a) [Cu(x)]` = C `Lim_(x->a) [u(x)]` = CP

Example 3:

Explanation:

` Lim_(x->3)(2x) = 2 Lim_(x->3)(x) = 2 (3) = 6 `

4. The limit of the Product of two functions is equal to the product of the limits of the two functions.

`Lim_(x->a) [u(x) v(x)] `= `[Lim_(x->a) u(x)] [ Lim_(x->a) v(x)]` = P Q

Example 4:

Explanation:

`Lim_(x->3)(2x)(x – 1)` = `[Lim_(x->3)(2x)][Lim_(x->3)(x – 1)] `= `[2(3)][3 – 1] `
= (6)(2) = 12

5. The limit of the quotient of two functions is equal to the quotient of the limits of the two functions (provided the limit of the denominator is non-zero).

`Lim_(x->a) (( u(x) )/( v(x)))` = ` ( Lim_(x->a) u(x) )/(Lim_(x->a) v(x))`= ` ( P )/( Q )(`Provided `Q ≠ 0)`

Example 5:

Explanation:

`Lim_(x->1) (( 2x+1 )/( x+2)) = ((Lim_(x->1) (2x+1) )/(Lim_(x->1) (x+2)))` = `( 2(1)+1 )/( 1+2)`= ` ( 3 )/( 3)= 1 `

6. The limit of any integral power of [f(x)]is equal to that power of the limit of `[f(x)]`.

`Lim_ (x->a) [ u(x)] ^n = (Lim_(x->a) [ u(x)] )^n = P^n `

Example 6:

Explanation:

`Lim_(x->4)` `(3x – 9)^4` = `(Lim_(x->4) (3x – 9) )^ 4` = `(3(4) – 9) ^4` = ` (3) ^4 = 81`

Limits of `(( 0 )/( 0 ))`form


In Calculus, many a times there comes a situation where putting the value of `x` gives us an expression of the form `( 0 )/( 0 )` .When such a situation arises, we use simplification method to factorize both the numerator and denominator expressions to see if there is a factor common to both. If such a common factor does occur, we cancel them about and are usually left with an expression in which putting the limit does not give us ` ( 0 )/( 0 ) ` form. We have presented some of the examples of such a scenario here:


Example 7:

Explanation:

`Lim_(x->1) ( x^2 - 1 )/(x^2- x )`= `((1)^2 - 1 )/((1)^2- 1 )`= ` ( 0 )/( 0 ) `

Hence we try to factorize the terms.


` Lim_(x->1) ( x^2 - 1 )/(x^2- x )= Lim_(x->1) ((x –1)(x+1) )/(x( x-1) )= Lim_(x->1)( (x+1) )/(x ) = ( (1+1) )/1 = 2 `

Limits at Infinity


In evaluating limits at infinity, we divide both the numerator and the denominator terms by the highest power of variable that appears in the denominator and then put the limit in the new formed expression.


Limits at Infinity Examples


Example 1:

Explanation:

We first divide both the numerator and the denominator by the highest power of x in the denominator which is 3.

`Lim_(x-> 8)((3x)^2/( x)^3 – ( 2x)/( x)^3 )/(( 4x)^3/( x)^3 - ( 2x)^2/( x)^3 + ( 2)/( x)^3 )=Lim_(x-> 8)(( 3)/( x ) – ( 2)/( x)^2 )/( 4 – ( 2 )/( x ) + ( 2)/( x)^3 )`

Putting the limit value of X = 8


` Lim_(x-> 8)(3x^2 – 2x )/( 4x^3- 2x^2+ 2 ) = (0 – 0 )/( 4 – 0 + 0 )= 0`

If ` theta ` is measured in radians, then `Lim_(T -> 0)( sin T)/T=1`


This is an important Mathematical result and will be used in many examples that you will come across in the subject of Limits and Calculus. We are not including the proof of this identity here as it might be out of your scope. We will focus more on understanding the solution strategies.
We solve a few examples that make use of this identity.


Example 2:

Explanation:

We are supposed to evaluate the

`Lim_(T -> 0)( sin 4T)/T` Putting ` x = 4T =>T= `x / 4

` Lim_(T -> 0) ( sin 4T)/T ` = `Lim_(x -> 0) ( sin x )/(x/4) ` `(->x -> 0 ` as` T -> 0)`

= 4 `Lim_(x -> 0) ( sin x )/x`

= 4 (1) = 4 ` (∵ Lim_(x -> 0)( sin x )/x = 1)`

Example 3:

Explanation:

If we directly put the value of limit x into the expression, we get a `( 0 )/( 0 ` form. If however we simplify the numerator term, it becomes easy to evaluate the limit as shown below.

`Lim_(x -> -1)( x^3 - x)/(x + 1 ) ` = `Lim_(x -> -1) ( x(x^2 – 1) )/(x + 1 )`= ` Lim_(x -> -1) (x(x – 1)(x +1))/(x + 1 )`

= `Lim_(x -> -1) x(x-1)` = (–1) (–1 – 1) = 2






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