# Precise Definition of a Limit

## Homework Statement

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

## The Attempt at a Solution

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

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

Last edited:

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Looks good.

Bacle2
Small nitpick on calculations:

in part 3, after "then" , at the end, the factors should be |x-5||x+5|.

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

$(\sqrt{x^2+16}\ )^2 - 5^2=x^2+16-25=x^2-9=\dots$

Bacle2