Prove that limit as x approaches three of x^2 is equal to 9

[WK95]

Homework Statement

Prove that
$\lim_{x \rightarrow 3} x^{2} = 9$

Homework Equations

For every ε>0, there is a δ<0 so that if 0<|x-a|<δ then |f(x)-L|<ε

The Attempt at a Solution

$If~0<|x - 3|<δ~then~|x^2 - 9|<ε$
$|x^2 - 9|$
$|x - 3||x + 3|$
$= |x - 3||x - 3 + 3 + 3|$
$= |x - 3|*|(x - 3) + 6|$
$≤ |x - 3|*(|x - 3| + |6|), by~triangle~inequality$
$= |x - 3|2 + 6|x - 3|$
$<$

 

[STEMucator]

Homework Statement

Prove that
$\lim_{x \rightarrow 3} x^{2} = 9$

Homework Equations

For every ε>0, there is a δ<0 so that if 0<|x-a|<δ then |f(x)-L|<ε

The Attempt at a Solution

$If~0<|x - 3|<δ~then~|x^2 - 9|<ε$
$|x^2 - 9|$
$|x - 3||x + 3|$

It was correct until there. Remember to use the fact that $|x-3| < \delta$ so you can write :

$|x - 3||x + 3| < \delta |x+3|$

Now apply the triangle inequality to $|x+3|$, what can you conclude?

 

[HallsofIvy]

Homework Statement

Prove that
$\lim_{x \rightarrow 3} x^{2} = 9$

Homework Equations

For every ε>0, there is a δ<0 so that if 0<|x-a|<δ then |f(x)-L|<ε

The Attempt at a Solution

$If~0<|x - 3|<δ~then~|x^2 - 9|<ε$
$|x^2 - 9|$
$|x - 3||x + 3|$

At this point, I would argue that if, say, |x- 3|< 1, then 2< x< 4 so that 5< x+ 3< 7.
If we want $|x- 3||x+ 3|< |x- 3|(7)< \epsilon$ then we must have
$|x- 3|< \frac{\epsilon}{7}$
So we can take $\delta$ to be the smaller of 1 and $\frac{\epsilon}{7}$

$= |x - 3||x - 3 + 3 + 3|$
$= |x - 3|*|(x - 3) + 6|$
$≤ |x - 3|*(|x - 3| + |6|), by~triangle~inequality$
$= |x - 3|2 + 6|x - 3|$
$<$