- #1
kylera
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As the subject says, I have "proved" a function, but because this type of problem seems to have no set answer, I'd like some opinions on whether I walked through it properly or not.
lim(x=> -2) (x^2-1) = 3
Precise Definition of a Limit
Part 1. Assume a value for [tex]\delta[/tex]
Since 0 < |x + 2|<[tex]\delta[/tex],
|f(x) - 3| < [tex]\epsilon[/tex]
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< [tex]\epsilon[/tex]
Let C = |x - 2|, which leads to C|x + 2| < [tex]\epsilon[/tex]
|x + 2| = [tex]\delta[/tex] < [tex]\frac{\epsilon}{C}[/tex]
Applying the Precise Definition of a Limit,
For a [tex]\delta[/tex] value [tex]\frac{\epsilon}{C}[/tex] greater than zero, there exists [tex]\epsilon[/tex] greater than zero such that if 0 < |x + 2|<[tex]\delta[/tex], then |f(x) - 3| < [tex]\epsilon[/tex].
|f(x) - 3| < [tex]\epsilon[/tex]
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< [tex]\epsilon[/tex]
Re-apply C and [tex]\delta[/tex] to get
C x [tex]\frac{\epsilon}{C}[/tex] < [tex]\epsilon[/tex]
Hence, by the Precise Definition of a limit, said limit does exist.
Much thanks in advance. Comments and criticisms are always welcome.
Homework Statement
lim(x=> -2) (x^2-1) = 3
Homework Equations
Precise Definition of a Limit
The Attempt at a Solution
Part 1. Assume a value for [tex]\delta[/tex]
Since 0 < |x + 2|<[tex]\delta[/tex],
|f(x) - 3| < [tex]\epsilon[/tex]
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< [tex]\epsilon[/tex]
Let C = |x - 2|, which leads to C|x + 2| < [tex]\epsilon[/tex]
|x + 2| = [tex]\delta[/tex] < [tex]\frac{\epsilon}{C}[/tex]
Applying the Precise Definition of a Limit,
For a [tex]\delta[/tex] value [tex]\frac{\epsilon}{C}[/tex] greater than zero, there exists [tex]\epsilon[/tex] greater than zero such that if 0 < |x + 2|<[tex]\delta[/tex], then |f(x) - 3| < [tex]\epsilon[/tex].
|f(x) - 3| < [tex]\epsilon[/tex]
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< [tex]\epsilon[/tex]
Re-apply C and [tex]\delta[/tex] to get
C x [tex]\frac{\epsilon}{C}[/tex] < [tex]\epsilon[/tex]
Hence, by the Precise Definition of a limit, said limit does exist.
Much thanks in advance. Comments and criticisms are always welcome.