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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!

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