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Precise Definition of a Limit

  • Thread starter drawar
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  • #1
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Homework Statement



Using the epsilon and delta definition, prove that:
[tex]\mathop {\lim }\limits_{x \to - 3} \sqrt {{x^2} + 16} = 5[/tex]

Homework Equations





The Attempt at a Solution



Given epsilon > 0. Choose [tex]\delta {\rm{ = min}}\left\{ {{\rm{1,}}\frac{{\left( {5 + \sqrt {20} } \right)\varepsilon }}{7}} \right\}[/tex], then:
[tex]0 < \left| {x + 3} \right| < \delta \Rightarrow \left| {\sqrt {{x^2} + 16} - 5} \right| = \frac{{\left| {x + 3} \right|.\left| {x - 3} \right|}}{{5 + \sqrt {{x^2} + 16} }}[/tex]
Moreover, [tex]\left| {x + 3} \right| < 1 \Rightarrow \left| {x - 3} \right| < 7[/tex] and [tex]5 + \sqrt {{x^2} + 16} > 5 + \sqrt {20} [/tex]
Hence, [tex]\frac{{\left| {x + 3} \right|.\left| {x - 3} \right|}}{{5 + \sqrt {{x^2} + 16} }} < \frac{{\frac{{\left( {5 + \sqrt {20} } \right)\varepsilon }}{7}.7}}{{5 + \sqrt {20} }} = \varepsilon .[/tex]This completes the proof.

Please correct if there's anything wrong with it. Thanks!
 
Last edited:

Answers and Replies

  • #2
6,054
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Looks good.
 
  • #3
Bacle2
Science Advisor
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Small nitpick on calculations:

in part 3, after "then" , at the end, the factors should be |x-5||x+5|.
 
  • #4
6,054
390
Small nitpick on calculations:

in part 3, after "then" , at the end, the factors should be |x-5||x+5|.
How did you get that?
 
  • #5
SammyS
Staff Emeritus
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Small nitpick on calculations:

in part 3, after "then" , at the end, the factors should be |x-5||x+5|.
You may want to recheck that.

[itex](\sqrt{x^2+16}\ )^2 - 5^2=x^2+16-25=x^2-9=\dots[/itex]
 
  • #6
Bacle2
Science Advisor
1,089
10
How did you get that?
Never mind, my bad. Computation mistake from jumping-in too quickly.
 

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