Concept of Limits - Limits of Polynomials and Rational Functions

A function f is said to be a polynomial function of degree n f(x) = ${\displaystyle a_0+a_{1}x+a_2{x}^2+..+a_n{x}^n}$  , where a_1s are real numbers such that ${\displaystyle a_n}$  ≠ 0 for some natural number n. 
${\displaystyle \underset{x\to a}{lim}}$ x = a .
Hence 
${\displaystyle \underset{x\to a}{lim}{x}^2=\underset{x\to a}{lim}(x.x)}$ = ${\displaystyle \underset{x\to a}{lim}x.\underset{x\to a}{lim}x=a.a=a^2}$

An easy exercise in induction on n tells us that
${\displaystyle \underset{x\to a}{lim}{x}^n=a^n}$
Now, let f(x) = ${\displaystyle a_0+a_{1}x+a_2{x}^2+...+a_n{x}^n}$ be a polynomial function.
Suppose of each of ${\displaystyle a_0,a_{1}x,a_2{x}^2,....,a_n{x}^n}$ as a function , we have  

${\displaystyle \underset{x\to a}{lim}f\left(x\right)=\underset{x\to a}{lim}[a_0+a_{1}x+a_2{x}^2+...+a_n{x}^n]}$

= ${\displaystyle \underset{x\to a}{lim}{a}_0+\underset{x\to a}{lim}{a}_{1}x +\underset{x\to a}{lim}{a}_2{x}^2+...+a_n{x}^n}$

= ${\displaystyle a_0+ a_1\underset{x\to a}{lim}x+a_2\underset{x\to a}{lim}{x}^2+...+a_n\underset{x\to a}{lim}{x}^n.}$

= ${\displaystyle a_0+a_{1}a+a_2{a}^2+...+a_n{a}^n}$

= f(a)
A function f is said to be a rational function, if f(x) = ${\displaystyle \frac{g\left(x\right)}{h\left(x\right)}}$ ,  where g(x) and h(x) are polynomials such that h(x) ≠ 0.
Then ${\displaystyle \underset{x\to a}{lim}f\left(x\right)=\underset{x\to a}{lim}\frac{g\left(x\right)}{h\left(x\right)}=\frac{\underset{x\to a}{lim}g\left(x\right)}{\underset{x\to a}{lim}h\left(x\right)}=\frac{g\left(a\right)}{h\left(a\right)}}$.

Case 1 - h(a) = 0  and g(a) = k

${\displaystyle \frac{g\left(a\right)}{h\left(a\right)}=\frac{k}{0}=\infty }$
Limit does not exist (undefined)
Example - 
${\displaystyle \underset{x\to 2}{lim}\frac{x^3-2}{x-2}=\frac{2^3-2}{2-2}=\frac{8-2}{0}=\frac{6}{0}=\infty }$

Case 2 -  
h(a) = 0 and  g(a) = 0

${\displaystyle \frac{g\left(a\right)}{h\left(a\right)}=\frac{0}{0}}$
Example - ${\displaystyle \underset{x\to 2}{lim}\frac{x^2-4}{x-2}=\frac{2^2-4}{2-2}=\frac{4-4}{2-2}=\frac{0}{0}}$

Case 3 -  h(a) = k  and g(a) = 0
${\displaystyle \frac{g\left(a\right)}{h\left(a\right)}=\frac{0}{k}=0}$
Example - ${\displaystyle \underset{x\to 2}{lim}\frac{x-2}{2x+2}=\frac{2-2}{2.2+2}=\frac{0}{4+2}=\frac{0}{6}=0}$