- #1

tomcenjerrym

- 37

- 0

[tex]lim_{x\rightarrow 2} (x^2 - 9) = -5[/tex]

And the solutions (by myself):

[tex]|f(x) - L| < \epsilon[/tex]

[tex]|(x^2 - 9) - (-5)| < \epsilon[/tex]

[tex]|x^2 - 9 + 5| < \epsilon[/tex]

[tex]|x^2 - 4| < \epsilon[/tex]

[tex]-\epsilon < x^2 - 4 < \epsilon[/tex]

[tex]-\epsilon + 4 < x^2 < \epsilon + 4[/tex]

It gets more complicated since the inequality in the QUADRATIC form.

Thus,

[tex]\sqrt{-\epsilon + 4} < \sqrt{x^2} < \sqrt{\epsilon + 4}[/tex]

[tex]\sqrt{-\epsilon + 4} < |x| < \sqrt{\epsilon + 4}[/tex]

Divide by 2 form:

**FIRST**

[tex]-\sqrt{\epsilon + 4} < x < \sqrt{\epsilon + 4}[/tex]

**SECOND**

[tex] x > \sqrt{-\epsilon + 4}[/tex] and [tex]x < -\sqrt{-\epsilon + 4}[/tex]

So, can anyone check my answer whether it’s true or not?