Mathematics LIMITS - BASICS AND DEFINITIONS

Topics Covered

`star` Introduction
`star` Right hand limit and Left hand limit
`star`Existence of Limit
`star` Algebra of limits
`star` Limits of polynomial functions
`star` Limits of rational functions
`star` For any positive integer `n,` `\ \ \ \ \ \ \ lim_(x→a) (x^n- a^n)/(x-a)= n a^(n-1)`

Introduction

`\color{green} ✍️` Consider the function `f(x) = x^2.` . Observe that as `x` takes values very close to `0,` the value of `f(x)` also moves towards `0`

`color(red)(lim_(x->0)f(x) = 0)` (`color(blue)("to be read as limit of f (x) as x tends to zero equals zero"`).

The limit of `f (x)` as `x` tends to zero is to be thought of as the value `f (x)` should assume at `x = 0.`

In general as `x → a, f (x) → l,` then `l` is called limit of the function `f (x)` which is symbolically written as `lim_(x->a)f(x) = l`.

Intuitive Idea of Derivatives

Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of `4.9t^2` metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by `s = 4.9t^2`.

The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff.

The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds.

Average velocity between `t = t_1` and `t = t_2` equals distance travelled between `t = t_1` and `t = t_2` seconds divided by (t2 – t1). Hence the average velocity in the first two seconds

`= ("Distance travelled between" t_2 = 2 "and" t_1 = 0)/("Time interva"l (t_2 - t_1) )`

`= ((19.6 - 0 ) m)/((2 -0) s) = 9.8` m/s

Similarly, the average velocity between t = 1 and t = 2 is

`(19.6 - 4.9 )m) /(( 2-1)s) = 14.7` m/s

Likewise we compute the average velocitiy between `t = t_1` and t = 2 for various `t_1`. The following

Table 13.2 gives the average velocity (v), `t = t_1` seconds and t = 2 seconds.

From Table 13.2, we observe that the average velocity is gradually increasing. As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551 m/s.

This conclusion is somewhat strengthened by the following set of computation. Compute the average velocities for various time intervals starting at t = 2 seconds. As before the average velocity v between t = 2 seconds and `t = t_2` seconds is

`= ("Distance travelled between 2 seconds and seconds" )/(t_2 -2)

`= ("Distance travelled in" `t_2` "seconds - Distance travelled in 2 seconds")/(t_2 -2)`

`= ("Distance travelled in seconds - 19.6")/(t_2 -2)`

The following Table 13.3 gives the average velocity v in metres per second between t = 2 seconds and `t_2` seconds.

Here again we note that if we take smaller time intervals starting at t = 2, we get better idea of the velocity at t = 2.

Right hand limit and Left hand limit

`color(red)(lim_(x->a)f(x) = l)`.

The value which the function should assume at a given point `x = a` did not really depend on how is `x` tending to `a.`

Note that there are essentially two ways `x` could approach a number a `color(blue)("either from left or from right")`, i.e., all the values of `x` near `a` could be less than `a` or could be greater than `a. `

This naturally leads to two limits – the `color(blue)(text(right hand limit))` and the `color(blue)(text(left hand limit.))`

Right hand limit of a function `f(x)` is that value of `f(x)` which is dictated by the values of `f(x)` when `x` tends to `a` from the right. Similarly, the left hand

Thus , `color(blue)(lim_(x->a^-)f(x) = l)` , if `f(x)` tends to `l` as `x` tends to `a` from the left-hand side.

`color(blue)(lim_(x->a^+) f(x) = l)` , if `f(x)` tends to `l'` as `x` tends to `a` from the right-hand side.

`color(red)(=> "Consider the function")`

`color(red)(f(x) = { tt (( 1, x le 0),(2, x > 0))`

Graph of this function is shown in the Fig 13.3.

It is clear that the value of `f` at `0` dictated by values of `f(x)` with `x ≤ 0` equals 1, i.e., the left hand limit of `f (x)` at `0` is

`lim_(x->0^(-)) f(x) =1`

Similarly, the value of f at 0 dictated by values of `f (x)` with `x > 0` equals `2`., i.e., the right hand limit of `f (x)`
at `0` is

`lim_(x->0^(+)) f(x) =2`

In this case the right and left hand limits are different, and hence we say that the limit of `f (x)` as x tends to zero does not exist (even though the function is defined at `0`).

Existence of Limit

`color(red)(lim_(x->a) f(x) )` exist if `color(blue)(lim_(x->a^-) f(x)) ` and `color(blue)(lim_(x->a^+)f(x)) ` exists and both are equal.

If the right and left hand limits coincide, we call that common value as the limit of `f(x)` at `x = a` and denote it by `color(blue)(lim_(x->a) f(x) )`

`lim_(x->a) f(x) = color(blue)(lim_(x ->a^-) f(x)= lim_(x->a^+)f(x).)` Then `color(blue)(lim_(x->a) f(x)) ` exist .

Q 3119780610

Consider the function `f(x) = x + 10`. We want to find the limit of this function at `x = 5`.

Solution:

Let us compute the value of the function `f(x)` for x very near to `5.`

Some of the points near and to the left of `5` are `4.9, 4.95, 4.99, 4.995. ... .,` etc. Values of the function at these points are tabulated below.

Similarly, the real number `5.001, 5.01, 5.1` are also points near and to the right of `5`. Value of the function at these points are also given in the Table 13.4.

From the Table 13.4, we deduce that value of `f(x)` at `x = 5` should be greater than `14.995` and less than `15.001 `

assuming nothing dramatic happens between `x = 4.995` and `5.001`. It is reasonable to assume that the value of the `f(x)` at `x = 5` as dictated by the numbers to the left of `5` is `15`, i.e.,

`lim_(x->5^(-)) f(x)=15`

Similarly, when x approaches `5` from the right, `f(x)` should be taking value `15`, i.e.,

`lim_(x->5^(+)) f(x)=15`

Hence, it is likely that the left hand limit of f(x) and the right hand limit of `f(x)` are both equal to 15. Thus,

`lim_(x->5^(-)) =lim_(x->5^(+)) f(x) =lim_(x->5) f(x) =15`

This conclusion about the limit being equal to `15`

In this figure, we note that as `x` appraches `5` from either right or left, the graph of the function

`f(x) = x +10` approaches the point `(5, 15).`

Q 3149780613

Consider the function `f(x) = x^3`. Find the limit of this function at `x = 1`.

Solution:

Proceeding as in the previous case, we tabulate the value of `f(x)` at `x` near 1. This is given in the Table `13.5`

From this table, we deduce that value of `f(x)` at `x = 1` should be greater than `0.997002999` and less than `1.003003001` assuming nothing dramatic happens between `x = 0.999` and `1.001.`

It is reasonable to assume that the value of the `f(x)` at `x = 1` as dictated by the numbers to the left of `1` is `1,` i.e.,

`lim_(x->1^(-)) f(x)=1`

Similarly, when `x` approaches 1 from the right, `f(x)` should be taking value `1.,` i.e.,

`lim_(x-> 1^(+)) f(x) =1`

Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to `1.` Thus,

`lim_(x->1^(-)) f(x)= lim_(x->1^(+)) f(x) =lim_(x->1) f(x)=1`

This conclusion about the limit being equal to `1`

we note that as x approaches 1 from either right or left, the graph of the function `f(x) = x^3` approaches the point `(1, 1)`.

We observe, again, that the value of the function at `x = 1` also happens to be equal to `1.`

Q 3109780618

Consider the function `f(x) = 3x`. Let us try to find the limit of this function at `x = 2`.

Solution:

The following Table 13.6 is now self-explanatory.

As before we observe that as `x` approaches `2` from either left or right, the value of `f(x)` seem to approach `6.`

We record this as

`lim_(x->2^(-)) f(x) = lim_(x->2^(+)) f(x)= lim_(x->2) f(x)=6`

Its graph shown in Fig 13.4 strengthens this fact

Here again we note that the value of the function at `x = 2` coincides with the limit at `x = 2`.

Q 3119780619

Consider the constant function `f(x) = 3`. Let us try to find its limit at `x = 2`.

Solution:

This function being the constant function takes the same value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence

`lim_(x->2^(-)) f(x) = lim_(x->2^(+)) f(x) = lim_(x->2) f(x) =3`

Graph of `f(x) = 3` is anyway the line parallel to `x`-axis passing through `(0, 3)`

From this also it is clear that the required limit is `3.` In fact,

it is easily observed that `lim_(x->a) f (x) =3` for any real number `a`.

Q 3119880710

Consider the function `f(x) = x^2 + x`. We want to find `lim_(x->1) f(x)`

Solution:

We tabulate the values of `f(x)` near `x = 1` in Table `13.7`

From this it is reasonable to deduce that

`lim_(x->1^(-)) f(x) =lim_(x->1^(+)) f(x) =lim_(x->1) f(x) =2`

From the graph of `f(x) = x^2 + x` shown in the Fig `13.5`,

it is clear that as `x` approaches `1,` the graph approaches `(1, 2).`

Here, again we observe that the

`lim_(x->1) f(x) =f(1)`

Now, convince yourself of the following three facts:

`lim_(x->1) x^2 =1, lim_(x->1) x=1 ` and `lim_(x->1) x+1=2`

Then `lim_(x->1) x^2 + lim_(x->1) x=1+1 =2 =lim_(x->1) [x^2 +x]`

Also, `lim_(x->1) x * lim_(x->1) (x+1) = 1* 2 = 2 =lim_(x->1) [ x (x+1)] = lim_(x->1) [ x^2+x]`

Q 3129880711

Consider the function `f(x) = sin x.` We are interested in `lim_(x-> pi/2) sin x`, where the angle is measured in radians.

Solution:

Here, we tabulate the (approximate) value of `f(x)` near `pi/2` (Table 13.8). From this, we may deduce that

`lim_(x-> (pi/2)^-) f(x) = lim_(x->(pi/2)^+) f(x) =lim_(x-> pi/2) f(x)=1`

Further, this is supported by the graph of `f(x) = sin x`

In this case too, we observe that `lim_(x-> pi/2) sin x = 1`.

The Algebra Of Limits

Let `f` and `g` be two functions such that both `lim_(x->a)f(x)` and `lim_(x->a)g(x)` exist. Then

` color(red)((i))` Limit of sum of two functions is sum of the limits of the functions, i.e.,

`color(blue) [lim_(x->a)[f(x) + g)(x) ] ]= color(blue){ lim_(x->a)f(x) + lim_(x->a) g(x)} `

` color(red)((ii))` Limit of difference of two functions is difference of the limits of the functions, i.e.,

`color(blue) [ lim_(x->a)[f(x) - g)(x)]] = color(blue){lim_(x->a)f(x) - lim_(x->a) g(x)} `

` color(red)((iii))` Limit of product of two functions is product of the limits of the functions, i.e.,

`color(blue) [lim_(x->a)[f(x)* g)(x)]]= color(blue){lim_(x->a)f(x) *lim_(x->a) g(x)} `

` color(red)((iv))` Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e.,

`color(blue){lim_(x->a)f(x)/g(x) = (lim_(x->a)f(x))/(lim_(x->a) g(x))}`

` color{red} {"Key Points"}` In particular as a special case of `(iii)`, when `g` is the constant function such that `g(x) = λ` , for some real number `λ` , we have

`color(blue)(lim_ (x->a) lamda*f(x) = lamda lim_(x->a)f(x)) `, where `lamda` is a constant.

In the next two subsections, we illustrate how to exploit this theorem to evaluate limits of special types of functions

Limits of polynomial functions

`\color{green} ✍️` A function `f` is said to be a polynomial function if `f(x)` is zero function or

if `color(red)(f(x) = a_0 + a_1x + a_2x^2 +. . . + a_nx^n,)`

where `a_i` are real numbers such that `a_n ≠ 0` for some natural number `n.`

We know that `color{blue}{lim_(x->a) x=a}`

Now, let `f(x) = a_0 +a_1x+a_2x^2 +.........+ a_nx^n` be a polynomial function. Thinking of each of `a_0 , a_1x , a_2x^2 , ........ , a_nx^n` as a function, we have

` =>color(blue){ lim_(x→a) f(x) = lim_(x→a) [a_0 +a_1x+a_2x^2+......+ a_nx^n]}`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= lim_(x→a) a_0 +lim_(x→a) a_1x+lim_(x→a) a_2x^2+.....+ lim_(x→a) a_nx^n`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= a_0 +a_1 lim_(x→a) x+a_2 lim_(x→a) x^2+.............+ a_n lim_(x→a) x^n`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a_0 +a_1 a+a_2 a^2+................+ a_n a^n`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color(blue)(= f(a))`

Limits of rational functions

`\color{green} ✍️` A function `f` is said to be a rational function, if `f(x) = g(x) / (h(x) ) , ` where `g(x)` and `h(x)` are polynomials such that `h(x) ≠ 0.` Then

`color{blue}{lim_(x->a) f(x) =lim_(x->a)g(x)/(h(x)) = (lim_(x->a)g(x))/(lim_(x->a) h(x)) =g(a)/(h(a))}`

However, if `h(a) = 0,` there are two scenarios –

`(i)` when `color{blue}{g(a) ≠ 0}` and

`(ii)` when `color{blue}{g(a) = 0.}` In the former case we say that the limit does not exist.

All the functions under consideration are rational functions.

Hence, we first evaluate these functions at the prescribed points. If this is of the form `0/0 ,` we try to rewrite the function cancelling the factors which are causing the limit to be of the form `0/0 .`

Q 3019745610

Find the limits: (i) `lim_(x→1) [ x^3-x^2+1]`
(ii). `lim_(x→3) [x (x+1)]`
(iii). `lim_(x→-1) [ 1+x+x^2+....+ x^(10) ]`

Solution:

The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have

(i). `lim_(→1) [x^3-x^2+1] = 1^3-1^2+1 = 1`

(ii). `lim_(x → 3) [ x(x+1) ] = 3 (3+1) = 3(4) = 12`

(iii). `lim_(x→ -1) [ 1+x+x^2+.......+ x^(10) ] = 1+(-1) + (-1)^2+.........+ (-1)^(10)`

` = 1-1+1...........+1 = 1`

Q 3039845712

Find the limits (i). `lim_(x→1) [ (x^2+1)/(x+100)]`
(ii). `lim_(x→2) [ ( x^3-4x^2+4x)/(x^2-4)]`
(iii). `lim_(x→2) [ (x^2-4)/(x^3-4x^2+4x)]`
(iv). `lim_(x→2) [ (x^3-2x^2)/(x^2-5x+6)]`
(v). `lim_(x→1) [ (x-2)/(x^2-x) - 1/(x^3-3x^2+2x)]`

Solution:

All the functions under consideration are rational functions. Hence, we first evaluate these functions at the prescribed points. If this is of the form `0/0` , we try to rewrite the function cancelling the factors which are causing the limit to be of the form `0/0`

(i). `lim_(x→1) (x^2+1)/(x+100) = (1^2+1)/(1+100) = 2/101`

(ii). Evaluating the function at 2, it is of the form `0/0` hence `lim_(x→2) (x^3-4x^2+4x)/(x^2-4) = lim_(x→2) (x(x-2)^2)/((x+2)(x-2))`

`= lim_(x→2) (x(x-2))/(( x+2)) ` as `x ne 2`

` = (2 (2-2))/(2+2) = 0/4 = 0`

(iii) Evaluating the function at 2, we get it of the form `0/0`.

Hence `lim_(x→2) (x^2-4)/(x^3-4x^2+4x) = lim_(x→2) ((x+2)(x-2))/(x(x-2)^2)`

` = lim_(x→2) ((x+2))/(x-2) = (2+2)/(2(2-2) = 4/0`

which is not defined. (iv) Evaluating the function at 2, we get it of the form `0/0`

Hence `lim_(x→2) (x^3-2x^2)/(x^2-5x+6) = lim_(x→2) (x^2 (x-2))/((x-2)(x-3))`

`= lim_(x→2) x^2/(x-3) = (2)^2/(2-3) = 4/(-1) = -4`

(v) First, we rewrite the function as a rational function.

`[ (x-2)/(x^2-x) - 1/(x^3-3x^2+2x)] = [(x-2)/(x(x-1)) - 1/(x(x^2-3x+2))]`

` = [ (x-2) /(x(x-1)) - 1/(x(x-1) (x-2))]`

`= (x^2-4x+3)/(x(x-1)(x-2))`

Evaluating the function at 1, we get it of the form `0/0`

Hence `lim_(x→1) [ (x^2-2)/(x^2-x) - 1/(x^3-3x^2+2x)] = lim_(x→1) (x^2-4x+3)/(x(x-1)(x-2))`

`= lim_(x→1) (x-3)/(x(x-2)) = (1-3)/(1(1-2)) = 2`

We remark that we could cancel the term (x – 1) in the above evaluation because x ≠ 1 .

For any positive integer `n,` `\ \ \ \ \ \ \ lim_(x→a) (x^n- a^n)/(x-a)= n a^(n-1)`

For any positive integer `n`

`color{red}{lim_(x→a) (x^n- a^n)/(x-a) = n a^(n-1)}`

`color{green}"Remark"` The expression in the above theorem for the limit is true even if `n` is any rational number and `a` is positive.

`color{green}"Proof : "` Dividing `(x^n – a^n)` by `(x – a)`, we see that

`x^n - a^n = (x-a) (x^(n-1) +x^(n-2)a +x^(n-3) a^2+.............+ x a^(n-2)+a^(n-1) )`

Thus `color(blue)(lim_(x→a) (x^n-a^n)/(x-a)) = lim_(x→a) (x^(n-1) +x^(n-2) a +x^(n-3) a^2+........+xa^(n-2) +a^(n-1))`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= a^(n-1) +a* a^(n-2) +.......+ a^(n-2) (a) +a^(n-1)`

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= a^(n-1) +a^(n-1) +...........+ a^(n-1) +a^(n-1) ` (n terms)

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(= n a^(n-1))`

Q 3089045817

Evaluate: (i). `lim_(x→1) (x^(15) -1)/(x^(10) -1)`

(ii). `lim_(x→0) (sqrt(1+x)-1)/x`

Solution:

(i) We have `lim_(x→1) (x^(15)-1)/(x^(10)-1) = lim_(x→1) [ (x^(15) -1)/(x-1) div (x^(10) -1)/(x-1)]`

` = lim_(x→1) [ (x^(15)-1)/(x-1)] div lim_(x→1) [ (x^(10)-1)/(x-1)]`

` = 15(1)^(14) div 10(1)^9` (by the theorem above)

` = 15 div 10 = 3/2`

(ii). Put `y = 1 + x`, so that `y→1` as `x→0.`

Then `lim_(x→0) (sqrt(1+x) -1)/x = lim_(y→1) (sqrty -1)/(y-1)`

` = lim_(y →1) (y^(1/2) -1^(1/2))/(y-1)`

` = 1/2 (1)^(1/2-1)` (by the remark above) = 1/2`

Class 11 Chapter 13 - LIMITS AND DERIVATIVES

LIMITS - BASICS AND DEFINITIONS