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

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Please see under.

The precursive segregation is a bit desire. If you regular shortness to learn the test, scroll down.

Preliminary segregation

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

By restriction,

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

for whole ##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 shortness to reach ##abs(underbrace(color(red)((x^2-7x+3)))_(color(red)(f(x)) )-underbrace(color(blue)((-7)))_color(blue)(L))## close than some attached ##epsilon## and we regulate (through our regulate of ##delta##) the magnitude of ##abs(x-underbrace(color(green)((2)))_color(green)(a))##

Look at the monstrosity we shortness to reach minute. Retranscribe this, looking for the monstrosity we regulate.

##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 monstrosity we regulate

We can reach ##abs(x-5)abs(x-2)< epsilon## by making ##abs(x-2) < (epsilon)/abs(x-5)##, BUT we deficiency a ##delta## that is recalcitrant of ##x##. Here's how we can fruit about that.

If we reach firm that the ##delta## we still cull is close than or correspondent to ##1##, then for whole ##x## after a while ##abs(x-2) < delta##, we earn enjoy ##abs(x-2) < 1##

which is penny if and simply if ##-1 < x-2 < 1 ##

which is penny if and simply if ##1 < x < 3##

which, is at-last equipollent to ##-4 < x-5 < -2##.

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

If we as-well reach firm that ##delta <= epsilon/4##, then we earn enjoy:

for all ##x## after a while ##abs(x-2) < delta## we enjoy ##abs((x-2)(x-5)) < delta * 4 <= epsilon/4 * 4 = epsilon##

So we earn cull ##delta = min{1, epsilon/4}##. (Any closeer ##delta## would as-well fruit.)

Now we deficiency to really transcribe up the test:

Proof

Given ##epsilon > 0##, cull ##delta = min{1, epsilon/4}##. ##" "## (hush that ##delta## is as-well actual).

Now for whole ##x## after a while ##0 < abs(x-2) < delta##, we enjoy

##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, after a while this excellent of delta, whenever ##0 < abs(x-2) < delta##, we enjoy ##abs((x^2-7x+3)-(-7)) < epsilon##

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


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