How do you use the limit definition to prove a limit exists?

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The formal definition of a limit is:
##lim_(x->c) f(x) = L## means that for every ##epsilon>0##, there exists a ##delta > 0## such that for every ##x## the expression ##0 < | x-c | < delta## implies ##|f(x) – L | < epsilon##

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We need to understand what this definition means.

“for every ##epsilon > 0## ” means that our proof must work for every ##epsilon##

“there exists a ##delta > 0## means our proof will give us a value for delta, normally depending upon the value of epsilon

“such that for every x” means we cannot restrict the values of x
“the expression ## 0 < |x-c| < delta## is the first step of the proof. The expression means that the values of x will be no more than delta units away from the value of c, where ## x != c##

“implies ##|f(x)-L| < epsilon ##” is the final step of the proof. Once we get to this, we have completed the proof.

Example:
Verify that the ##lim_(x->2) x^2 = 4##
Here ##a = 2 and L = 4##
Let ##epsilon > 0##. We need to prove that there exists a ##delta > 0## such that:
If ##0<|x-2 | < delta##, then ## | f(x) – 4 | < epsilon ##
## |f(x) -4 | = | x^2 -4 | = | (x+2)(x-2)| = | x + 2||x-2|##

We want this expression to be less than ##epsilon##
We need to ensure ##|x+2|## does not become too large.

If we assume that ##delta <= 1## and require ##|x-2 | < delta ##
## | x-2 | < 1 => 1 < x < 3 => 3 < x +2 < 5 => | x + 2| < 5##
Hence:
## |f(x) – 4 |< 5|x-2|## if ## | x-2 | < delta <= 1##
But if ## 5 |x-2| < epsilon ## if ## |x-2| < epsilon/5 ## then we can take ##delta = min{1,epsilon/5}##, then

## | f(x) – 4 | < 5|x-2| < 5 * epsilon /5 = epsilon ## if ## | x-2| < delta##

This proves that ##lim_(x->2) f(x) = 4##

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