Hi All,

I was going through basic definition of limts wherein it is defined that

Limit of a function f(x) is said to be L iff,

|f(x) - L| > 0 whenever |x-a|>0
Source:
http://web.mit.edu/wwmath/calculus/limits/formal.html

meaning when 'x' is not 'a' then limit of f(x) at 'a' and value of f(x) at 'x' are also not equal.

consider the following example:
f(x) = x/x

this function is undefine at x=0;
Limt f(x) = 1
x->0
here a = 0


checking the above definition,
consider x = 1 , x!= a hence |x-a| >0
f(1) = 1 = L hence |f(x) - L|= 0 even if |x-a| >0

This is confusing me. Please guide.Please Rectify if I went wrong somewhere.

Take another look at the definition - there is nothing that requires |f(x)-L| > 0. What is says is that
\lim _{x \to a} f(x) = L
if for any positive number \epsilon there is a corresponding positive number \delta such that |f(x)-L| < \epsilon whenever 0 < |x-a| < \delta . So for your example of f(x) = 1, the limit as x \to 1 is indeed 1, because |f(x)-L| is always zero, and hence is always less than any positive \epsilon you care to name. Consider the case of \epsilon = 0.1 : then you're looking for a \delta such that |f(x) - 1| = |1 - 1| < 0.1. For example you could choose \delta = 0.5, meaning 0.5 < x < 1.5, you can then see that |f(x)-L| is less than \epsilon = 0.1 for that full range of x. Thus the limit is L = 1.

ebaines done a fine job explaining. But if I may suggest something. YouTube is loaded with math tutorials of all sorts.

Go here and watch the video:
YouTube - Epsilon Delta Limit Definition 1 (http://www.youtube.com/watch?v=-ejyeII0i5c)

Thanks ebaines, I looked the concept as
|f(x) - L|> Epsilon instead of
|f(x) - L|< Epsilon