Concepts

13 min read

- Let's dig deep into the concepts of the chapter

1

Let $lim_{x→a}f(x)=l$ .When we approach the point $x=a$ from the values which are greater than or smaller than $x=a$ ,$f(x)$ would have tendency to move closer to the value $l$.

2

We know that $anynumber0 =0$

Also, $0anynumber =∞$

Lets consider a very interesting case , $00 $

This case is absurd and can be investigated by this odd argument

If $a×d=b×c$ then clearly $ba =dc $

So if $6×0=11×0$ then clearly $116 =00 $

If we choose any two numbers instead of $6$ and $11$ we will get the same answer.

For eg:

Lets take $5$ and $12$ instead off $6$ and $11$.

$∴5×0=12×0$

$∴125 =00 $

This brings us to the conclusion that $00 $ can be equal to any number

The above conclusion is absurd.

Hence,we call $00 $ as**Indeterminate Form**

Also, $0anynumber =∞$

Lets consider a very interesting case , $00 $

This case is absurd and can be investigated by this odd argument

If $a×d=b×c$ then clearly $ba =dc $

So if $6×0=11×0$ then clearly $116 =00 $

If we choose any two numbers instead of $6$ and $11$ we will get the same answer.

For eg:

Lets take $5$ and $12$ instead off $6$ and $11$.

$∴5×0=12×0$

$∴125 =00 $

This brings us to the conclusion that $00 $ can be equal to any number

The above conclusion is absurd.

Hence,we call $00 $ as

3

Lets understand the concept of limits with the help of a function,

$f(x)=x−1x_{2}−1 $

Let's work it out for $x=1$:

$f(1)=00 $

$f(x)=x−1x_{2}−1 $

Let's work it out for $x=1$:

$f(1)=00 $

Now $00 $ is a difficulty! We don't really know the value of $00 $ since,it is indeterminate, so we need another way of answering this.

So instead of trying to find value of $f(x)$ at $x=1$ let's try **approaching **$x$ closer and closer to $1$ as shown.

$x$ | $0.5$ | $0.9$ | $0.99$ | $0.999$ | $0.9999$ | $0.99999$ |

$f(x)=x−1x_{2}−1 $ | $1.5$ | $1.9$ | $1.99$ | $1.999$ | $1.9999$ | $1.99999$ |

We are now faced with an interesting situation:

- When $x=1$ we don't know the value of $f(x)$ (it is
**indeterminate**) - But we can see that it is approaching
**the value 2**when $x$ is approaching $1$

We summarize the behaviour of $f(x)$ by using special word "limit"

We say that the **limit** of $f(x)$, as $x$ approaches $1$ is** **$2$

And it is written in symbols as:

$x→1lim f(x)=2$

since, $f(x)=x−1x_{2}−1 $ we can also represent it as,

$x→1lim x−1x_{2}−1 =2$

So it is a special way of saying,* "ignoring what happens to the function when $x=1$ , but as we get closer and closer to $1$ the value of the function gets closer and closer to $2$*

4

1.Limit of sum of two functions is the sum of the limits of the function.

$x→alim [f(x)+g(x)]=x→alim f(x)+x→alim g(x)$

2.Limit of difference of two functions is difference of the limits of the function.

$x→alim [f(x)−g(x)]=x→alim f(x)−x→alim g(x)$

3. Limit of product of two functions is product of the limits of the function.

$x→alim [f(x)⋅g(x)]=x→alim f(x)⋅x→alim g(x)$

4. Limit of quotient of two functions is quotient of the limits of the function. (if denominator is non zero).

$x→alim g(x)f(x) =x→alim g(x)x→alim f(x) $

(if $g(x)=0$)

$x→alim [f(x)+g(x)]=x→alim f(x)+x→alim g(x)$

2.Limit of difference of two functions is difference of the limits of the function.

$x→alim [f(x)−g(x)]=x→alim f(x)−x→alim g(x)$

3. Limit of product of two functions is product of the limits of the function.

$x→alim [f(x)⋅g(x)]=x→alim f(x)⋅x→alim g(x)$

4. Limit of quotient of two functions is quotient of the limits of the function. (if denominator is non zero).

$x→alim g(x)f(x) =x→alim g(x)x→alim f(x) $

(if $g(x)=0$)

5

Evaluate $x→1lim (x_{3}−x_{2}+1)$

Solution:

we first put $x=1$ in $x_{3}−x_{2}+1$

$∴x→1lim (x_{3}−x_{2}+1)=1−1+1$

$∴x→1lim (x_{3}−x_{2}+1)=1$

Solution:

we first put $x=1$ in $x_{3}−x_{2}+1$

$∴x→1lim (x_{3}−x_{2}+1)=1−1+1$

$∴x→1lim (x_{3}−x_{2}+1)=1$

6

Any Function of the type $q(x)p(x) $ (where $p(x)$ and $q(x)$ are polynomials, and $q(x)=0$ ) is known as Rational Function.

To Solve Limits of a rational function.$x→alim q(x)p(x) $

To Solve Limits of a rational function.$x→alim q(x)p(x) $

- Put $x=a$ in $q(x)p(x) $, if we get $00 $ then proceed to next step.Else find the answer by direct substitution.
- Factorize $p(x)$ and $q(x)$ and cancel out the non zero common terms.
- Repeat Step 1 and Step 2 untill we get a real number as an answer.

7

Evaluate $x→0lim 2(sinx+2) $

Solution:

We first put $x=0$ in $2(sinx+2) $

$∴x→0lim 2(sinx+2) =20+2 $

$=1$

Solution:

We first put $x=0$ in $2(sinx+2) $

$∴x→0lim 2(sinx+2) =20+2 $

$=1$

8

If $f(x)$ ,$g(x)$ and $h(x)$ are three functions such that $f(x)≤g(x)≤h(x)$ for all $x$ in some interval containing the point $x=c$

such that $x→clim f(x)=x→clim h(x)=l$

then,$x→clim g(x)=l$

such that $x→clim f(x)=x→clim h(x)=l$

then,$x→clim g(x)=l$

9

If $f(x)$ and $g(x)$ are both derivable at $x=a$ , $f(x)±g(x)$ ,$f(x).g(x)$ and $g(x)f(x) $ will also be derivable at $x=a$.

If both $f(x)$ and $g(x)$ are non derivable , then nothing can be said about the sum /difference /product function.

if $f(x)$ is derivable at $x=a$ and $f(a)=0$ and $g(x)$ is continuous at $x=a$

If both $f(x)$ and $g(x)$ are non derivable , then nothing can be said about the sum /difference /product function.

if $f(x)$ is derivable at $x=a$ and $f(a)=0$ and $g(x)$ is continuous at $x=a$

10

$dxd(sinx) .$

$y=sinx=f(x),$(say)

$y=δy=sin(x+δx)=f(x+δx)$

$∴δy=(x+δx)−sinx=f(x+δx)−f(x)$

$∴δxδy =δxsin(x+δx)−sinx =δxf(x+δx)−f(x) $

$∴δxδy =δx→0lim δxsin(x+δx)−sinx $

$=lim_{δx→0}=δxf(x+δx)−f(x) $

$=δx→0lim =δx2cos(x+21 δx)sin21 δx $

$=δx→0lim cos(x+2δx )(δx/2)sin(δx/2) $

$=cosx.1=cosx(∵δx→0lim θsinθ =1.)$

$y=sinx=f(x),$(say)

$y=δy=sin(x+δx)=f(x+δx)$

$∴δy=(x+δx)−sinx=f(x+δx)−f(x)$

$∴δxδy =δxsin(x+δx)−sinx =δxf(x+δx)−f(x) $

$∴δxδy =δx→0lim δxsin(x+δx)−sinx $

$=lim_{δx→0}=δxf(x+δx)−f(x) $

$=δx→0lim =δx2cos(x+21 δx)sin21 δx $

$=δx→0lim cos(x+2δx )(δx/2)sin(δx/2) $

$=cosx.1=cosx(∵δx→0lim θsinθ =1.)$

11

If $f(x)$ and $g(x)$ are both derivable at $x=a$ , $f(x)±g(x)$ , $f(x).g(x)$

and $g(x)f(x) $ will also be derivable at $x=a$

and $g(x)f(x) $ will also be derivable at $x=a$

12

$u$ and $v$ are functions of $x$

$dxd (uv)=udxdv +vdxdv $

$dxd (uv)=udxdv +vdxdv $

13

Given functions of $g(x)$ and $h(x)$

$f(x)=h(x)g(x) $ and $h(x)=0$

$f_{′}(x)=(h(x))_{2}h(x)g_{′}(x)−g(x)h_{′}(x) $

$f(x)=h(x)g(x) $ and $h(x)=0$

$f_{′}(x)=(h(x))_{2}h(x)g_{′}(x)−g(x)h_{′}(x) $