Confirm Precise Definition of a Limit solution

[kylera]

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.

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 \delta
Since 0 < |x + 2|<\delta,

|f(x) - 3| < \epsilon
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< \epsilon

Let C = |x - 2|, which leads to C|x + 2| < \epsilon

|x + 2| = \delta < \frac{\epsilon}{C}

Applying the Precise Definition of a Limit,

For a \delta value \frac{\epsilon}{C} greater than zero, there exists \epsilon greater than zero such that if 0 < |x + 2|<\delta, then |f(x) - 3| < \epsilon.

|f(x) - 3| < \epsilon
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< \epsilon

Re-apply C and \delta to get

C x \frac{\epsilon}{C} < \epsilon

Hence, by the Precise Definition of a limit, said limit does exist.

Much thanks in advance. Comments and criticisms are always welcome.

 

[durt]

It looks like you're doing this backwards. Your proof should start like "Given \epsilon &gt; 0 there exists \delta &gt; 0 such that if 0&lt;|x+2|&lt;\delta, then |f(x)-3|&lt;\epsilon..." Now show that this is true using the $\delta$ that you found.

 

[HallsofIvy]

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.

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 \delta
Since 0 < |x + 2|<\delta,

|f(x) - 3| < \epsilon
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< \epsilon

Let C = |x - 2|, which leads to C|x + 2| < \epsilon

You can't do that. |x-2| is a variable, not a constant
what you need is a BOUND on |x-2|. If C is close to -2, say between -1 and -3, how large or how small can |x-2| be?

|x + 2| = \delta < \frac{\epsilon}{C}

Applying the Precise Definition of a Limit,

For a \delta value \frac{\epsilon}{C} greater than zero, there exists \epsilon greater than zero such that if 0 < |x + 2|<\delta, then |f(x) - 3| < \epsilon.

|f(x) - 3| < \epsilon
|f(x) - 3| = |x^2 - 1 - 3|
= |x^2 - 4| = |x + 2||x - 2|< \epsilon

Re-apply C and \delta to get

C x \frac{\epsilon}{C} < \epsilon

Hence, by the Precise Definition of a limit, said limit does exist.

Much thanks in advance. Comments and criticisms are always welcome.