Easy Way to Learn Limits

Introduction to Easy way to learn limits:      
In the easy way to learn limits, let f be a function of a real variable x. Let c and l be two unchanging numbers. If f(x)  come within reach of the value l as x approaches c, we say l is the limit of the function f(x) as x tends to c. This is written as

`lim_(x->c)`   f(x)   =  l.

Left Hand and Right Hand Limits:

In thr easy way to learn limits, although defining the limit of a function as x be inclined to c, we consider values of f(x) when x is very close to c (x>c or x
Lf(c) = `lim_(x->c )` _f(x) , provided the limit exists.

Likewise if x gets merely values greater than c, next x is said to tend to c from above or from right, and is denoted symbolically as x ? c + 0 or x ? c+, here the limit of f is then labeled the right hand limit. This is written as Rf(c) = `lim_(x->c+)` f(x). It is significant to reminder that for the survival of`lim_(x->c)` f(x) it is necessary that both Lf(c) and Rf(c) exists and Lf(c) = Rf(c) =`lim_(x->c)` f(x).  As well as the left and right hand limits are labeled as one sided limits.

Fundamental Results for Easy Way to Learn Limits:
The following rules for easy way to learn limits,

(1) If f(x) = k for all x, then
`lim_(x->c) `  f(x) = k.
(2) If f(x) = x for all x, then                                                                                                               
`lim_(x->c)`  f(x) = c.
(3) If f and g are two

functions having limits and k is a invariable then

(i)  `lim_(x->c)` k f(x) = k `lim_(x->c)` f(x)

(ii)  `lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) + `lim_(x->c)`  g(x)

(iii)  `lim_(x->c)`  [f(x) - g(x)] = `lim_(x-> c)`  f(x) - `lim_(x->c)`  g(x)

(iv)  `lim_(x->c)` c [f(x) . g(x)] = `lim_(x->c)` f(x) . `lim_(x->c)` g(x)

(v)    `lim_(x->c)` [`f(x)/g(x)` ] = `lim_(x->c)`  f(x)  / `lim_(x->c)`  g(x)              g(x) `!=` 0

(vi)  If   f(x) = g(x) then

`lim_(x-> c)`  f(x) =  `lim_(x->c)` g(x).

Example of Easy Way to Learn Limits:

Evaluate  `lim_(x->3)` `(x^2+ 7x + 11) / (x^2-9)` .

Solution:
Let f(x) = `(x^2 + 7x + 11)/( x^2-9)`
This is of the form f(x) =`g(x) / (h(x))`  ,
where g(x) = x2 + 7x + 11 and h(x) = x2 - 9. Clearly g(3) = 41 ? 0 and h(3) = 0.
Therefore f(3) = `g(3) /(h(3))`  = `41/ 0`  . Hence `lim_(x->3)`  `(x^2+ 7x + 11) / (x^2-9)` does not exist.