How do you use the epsilon delta definition to prove that the limit of ##x^2-7x+3=-7## as ##x->2##?

Please see below.

The preliminary analysis is a bit long. If you just want to read the proof, scroll down.

Preliminary analysis

We want to show that ##lim_(xrarr2)(x^2-7x+3) = -7##.

By definition,

##lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)## if and only if

for every ##epsilon > 0##, there is a ##delta > 0## such that:for all ##x##, ##” “## if ##0 < abs(x-color(green)(a)) < delta##, then ##abs(color(red)(f(x))-color(blue)(L)) < epsilon##.

So we want to make ##abs(underbrace(color(red)((x^2-7x+3)))_(color(red)(f(x)) )-underbrace(color(blue)((-7)))_color(blue)(L))## less than some given ##epsilon## and we control (through our control of ##delta##) the size of ##abs(x-underbrace(color(green)((2)))_color(green)(a))##

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

##abs((x^2-7x+3)-(-7)) = abs(x^2-7x+10)##

## = abs((x-5)(x-2))##

## = abs(x-5)abs(x-2)##

And there’s ##abs(x-2)##, the thing we control

We can make ##abs(x-5)abs(x-2)< epsilon## by making ##abs(x-2) < (epsilon)/abs(x-5)##, BUT we need a ##delta## that is independent of ##x##. Here’s how we can work around that.

If we make sure that the ##delta## we eventually choose is less than or equal to ##1##, then for every ##x## with ##abs(x-2) < delta##, we will have ##abs(x-2) < 1##

which is true if and only if ##-1 < x-2 < 1 ##

which is true if and only if ##1 < x < 3##

which, is ultimately equivalent to ##-4 < x-5 < -2##.

Consequently: if ##abs(x-2) < 1##, then ##abs(x-5) < 4##

If we also make sure that ##delta <= epsilon/4##, then we will have:

for all ##x## with ##abs(x-2) < delta## we have ##abs((x-2)(x-5)) < delta * 4 <= epsilon/4 * 4 = epsilon##

So we will choose ##delta = min{1, epsilon/4}##. (Any lesser ##delta## would also work.)

Now we need to actually write up the proof:

Proof

Given ##epsilon > 0##, choose ##delta = min{1, epsilon/4}##. ##” “## (note that ##delta## is also positive).

Now for every ##x## with ##0 < abs(x-2) < delta##, we have

##abs (x-5) < 4## and ##abs(x-2) < epsilon/4##. So,

##abs((x^2-7x+3)-(-7)) = abs(x^2-7x+10)##

## = abs(x-5)abs(x-2)##

## < 4 * delta <= 4 * epsilon/4 = epsilon##

Therefore, with this choice of delta, whenever ##0 < abs(x-2) < delta##, we have ##abs((x^2-7x+3)-(-7)) < epsilon##

So, by the definition of limit, ##lim_(xrarr2)(x^2-7x+3) = -7##.