Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I had the following question of a student this day about proving the following limit:

[tex]

\lim_{x \rightarrow 3} x^2 = 9

[/tex]

So this means that I should prove that

[tex]

|x-3| < \delta \ \rightarrow \ \ |x^2 - 9| < \epsilon(\delta)

[/tex]

So I had the following idea:

[tex]

|x^2 - 9| = |x-3||x+3|

[/tex]

The first term on the RHS is smaller than delta. For the second term I write

[tex]

|x+3| = |x-3+6| < |x-3| + 6

[/tex]

So I get in total

[tex]

|x-3| < \delta \ \rightarrow \ |x^2 - 9| = |x-3||x+3| < |x-3|(|x-3| + 6 ) < \delta(\delta + 6)

[/tex]

So choosing

[tex]

\epsilon = \delta(\delta + 6)

[/tex]

should prove the statement, right? But if I see the book they're using (Apostol) I see that they "choose" |x-3|<C, try to make an inequality then for |x+3| etc. But for me that seems making things more difficult then they are. So my question is: is my answer described above right?

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Delta-epsilon method

Loading...

Similar Threads - Delta epsilon method | Date |
---|---|

I Epsilon-delta vs. infinitesimal | Jan 28, 2017 |

I Help With Epsilon Delta Proof of multivariable limit | Jun 19, 2016 |

B Definition of limit problem | Jun 14, 2016 |

I Why non-standard analysis is not used? | Apr 1, 2016 |

I Epsilon-Delta definition property. | Mar 22, 2016 |

**Physics Forums - The Fusion of Science and Community**