Let ${\lim }_{x\to a}f(x)=l$ .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 $l$.

We know that $\frac{0}{anynumber}=0$ 
Also, $\frac{anynumber}{0}=\infty$
Lets consider a very interesting case , $\frac{0}{0}$
This case is absurd and  can be investigated by this odd argument
If $a\times d=b\times c$ then clearly $\frac{a}{b}=\frac{c}{d}$
So if $6\times 0=11\times 0$ then clearly $\frac{6}{11}=\frac{0}{0}$
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 $11$.
$\therefore 5\times 0=12\times 0$
$\therefore \frac{5}{12}=\frac{0}{0}$
This brings us to the conclusion that $\frac{0}{0}$ can be equal to any number
The above conclusion is absurd.
Hence,we call $\frac{0}{0}$ as Indeterminate Form

Lets understand the concept of limits with the help of a function,
$f(x)=\frac{x^2-1}{x-1}$
Let's work it out for $x=1$:
$f(1)=\frac{0}{0}$

Now $\frac{0}{0}$ is a difficulty! We don't really know the value of $\frac{0}{0}$ 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.

$x$	$0.5$	$0.9$	$0.99$	$0.999$	$0.9999$	$0.99999$
$f(x)=\frac{x^2-1}{x-1}$	$1.5$	$1.9$	$1.99$	$1.999$	$1.9999$	$1.99999$

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:

                          $\underset{x\to 1}{\lim }f(x)=2$

   since,    $f(x)=\frac{x^2-1}{x-1}$ we can also represent it as,

                          $\underset{x\to 1}{\lim }\frac{x^2-1}{x-1}=2$

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 $2$

1.Limit of sum of two functions is the sum of the limits of the function.
$\underset{x\to a}{\lim }[f(x)+g(x)]=\underset{x\to a}{\lim }f(x)+\underset{x\to a}{\lim }g(x)$
2.Limit of difference of two functions is difference of the limits of the function.
$\underset{x\to a}{\lim }[f(x)-g(x)]=\underset{x\to a}{\lim }f(x)-\underset{x\to a}{\lim }g(x)$
3. Limit of product of two functions is product of the limits of the function.
$\underset{x\to a}{\lim }[f(x)\cdot g(x)]=\underset{x\to a}{\lim }f(x)\cdot \underset{x\to a}{\lim }g(x)$
4. Limit of quotient of two functions is quotient of the limits of the function. (if denominator is non zero).
$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\frac{\underset{x\to a}{\lim }f(x)}{\underset{x\to a}{\lim }g(x)}$
(if $g(x)\ne 0$)

Evaluate $\underset{x\to 1}{\lim }(x^3-x^2+1)$
Solution:
we first put $x=1$ in $x^3-x^2+1$
$\therefore \underset{x\to 1}{\lim }(x^3-x^2+1)=1-1+1$
$\therefore \underset{x\to 1}{\lim }(x^3-x^2+1)=1$

Any Function of the type $\frac{p(x)}{q(x)}$ (where $p(x)$ and $q(x)$ are polynomials, and $q(x)\ne 0$ ) is known as Rational Function.
To Solve Limits of a rational function.$\underset{x\to a}{\lim }\frac{p(x)}{q(x)}$

1. Put $x=a$ in $\frac{p(x)}{q(x)}$, if we get $\frac{0}{0}$ then proceed to next step.Else find the answer by direct substitution.
2. Factorize $p(x)$ and $q(x)$ and cancel out the non zero common terms.
3. Repeat Step 1 and Step 2 untill we get a real number as an answer.

Evaluate $\underset{x\to 0}{\lim }\frac{(sinx+2)}{2}$
Solution:
We first put $x=0$ in $\frac{(sinx+2)}{2}$
$\therefore \underset{x\to 0}{\lim }\frac{(sinx+2)}{2}=\frac{0+2}{2}$
$=1$

If $f(x)$ ,$g(x)$ and $h(x)$ are three functions such that $f(x)\le g(x)\le h(x)$ for all $x$ in some interval containing the point $x=c$  
such that $\underset{x\to c}{\lim }f(x)=\underset{x\to c}{\lim }h(x)=l$ 
then,$\underset{x\to c}{\lim }g(x)=l$

If $f(x)$ and $g(x)$ are both derivable at $x=a$ , $f(x)\pm g(x)$ ,$f(x).g(x)$ and $\frac{f(x)}{g(x)}$ 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 $x=a$

$\frac{d(\sin x)}{dx}.$
$y=\sin x=f\left(x\right),$(say)
$y=\delta y=\sin \left(x+\delta x\right)=f\left(x+\delta x\right)$
$\therefore \delta y=\left(x+\delta x\right)-\sin x=f\left(x+\delta x\right)-f(x)$
$\therefore \frac{\delta y}{\delta x}=\frac{\sin \left(x+\delta x\right)-\sin x}{\delta x}=\frac{f\left(x+\delta x\right)-f\left(x\right)}{\delta x}$
$\therefore \frac{\delta y}{\delta x}=\underset{\delta x\to 0}{\lim }\frac{\sin \left(x+\delta x\right)-\sin x}{\delta x}$
$=li{m}_{\delta x\to 0}=\frac{f\left(x+\delta x\right)-f\left(x\right)}{\delta x}$
$=\underset{\delta x\to 0}{\lim }=\frac{2\cos \left(x+\frac{1}{2}\delta x\right)\sin \frac{1}{2}\delta x}{\delta x}$
$=\underset{\delta x\to 0}{\lim }\cos \left(x+\frac{\delta x}{2}\right)\frac{\sin \left(\delta x/2\right)}{\left(\delta x/2\right)}$
$=\cos x.1=\cos x\left(\because \underset{\delta x\to 0}{\lim }\frac{\sin \theta }{\theta }=1.\right)$

If $f(x)$ and $g(x)$ are both derivable at $x=a$ , $f(x)\pm g(x)$ , $f(x).g(x)$
and $\frac{f(x)}{g(x)}$ will also be derivable at $x=a$

$u$ and $v$ are functions of $x$
$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{dv}{dx}$

Given functions of $g(x)$ and $h(x)$
$f(x)=\frac{g(x)}{h(x)}$ and $h(x)\ne 0$
$f^{\prime }(x)=\frac{h(x)g^{\prime }(x)-g(x)h^{\prime }(x)}{{(h(x))}^2}$