Solving a \lim Problem with an \epsilon-\delta Argument

[ultima9999]

Hey, I'm stuck on this problem and just wanted some help if possible.

Prove, using an \epsilon - \delta argument, that \lim_{\substack{x\rightarrow -2}} \frac {1 - x - 2x^2} {x + 1} = 5

Ok, so I've answered so far with:

\lim_{\substack{x\rightarrow -2}} \frac {1 - x - 2x^2} {x + 1} = 5 if \forall \epsilon > 0, \exists \delta > 0 : 0 < |x + 2| < \delta \Rightarrow |-2x - 4| < \epsilon

Investigation to find \delta
|-2x - 4| < \epsilon
\Rightarrow -2|x + 2| < \epsilon
\Rightarrow |x + 2| > -\frac {\epsilon} {2}

Therefore, choose \delta = -\frac {\epsilon} {2}

My problem is here, my statement was that |x + 2| < \delta, but my answer so far has given me |x + 2| > \delta. Also, \delta is < 0.

What should I change?

 

[LeonhardEuler]

Here's where you made the mistake:

|-2x - 4| < \epsilon
\Rightarrow -2|x + 2| < \epsilon

Think about it: |-2x - 4| must be greater than or equal to zero, while -2|x+2| must be less than or equal to zero. They can not possibly be equal for all x.

 

[ultima9999]

I don't really understand what you're getting at...

edit: Could I do this?
\begin{align*}<br /> |-2x - 4| &lt; \epsilon\\<br /> \Rightarrow -2|x + 2| &lt; \epsilon\\<br /> \Rightarrow |x + 2| &gt; -\frac {\epsilon} {2}\\<br /> \Leftrightarrow |x + 2| &lt; \frac {\epsilon} {2} \ or\ |x + 2| &gt; -\frac {\epsilon} {2}<br /> \end{align*}

Therefore, choose \delta = \frac {\epsilon} {2}

Proof:
Let \epsilon &gt; 0 and set \delta = \frac {\epsilon} {2}

\begin{align*}<br /> |x + 2| &lt; \delta\\<br /> \Rightarrow |x + 2| &lt; \frac {\epsilon} {2}\\<br /> \Leftrightarrow -\frac {\epsilon} {2} &lt; |x + 2| &lt; \frac {\epsilon} {2}\\<br /> \Rightarrow |x + 2| &gt; -\frac {\epsilon} {2}\\<br /> \Rightarrow -2|x + 2| &lt; \epsilon\\<br /> \Rightarrow |-2x - 4| &lt; \epsilon<br /> \end{align*}
As required.

 

[matt grime]

The point is that you should not pull out negative numbers from inside an abs value symbol and leave them as negative: |-2x-4| = 2|x+2|.

It is 'vacuous' to say that -2|x+2|<e in the sense that it is always true since the left hand side is always negative and the right hand side is positive by assumption.

In effect you have gone through several unnecessary and dubious steps to conclude that |x+2|<e/2 which you could have done straight away by noting that |-2x-4|=2|x+2|. I say dubious because you have concluded that either this holds OR that |x+2|<e/-2, which is always true (for positive e) hence you have not got any restriction on x at all.

 

[ultima9999]

Ok, I fixed it up to make it look like this:Prove, using an \epsilon - \delta argument, that \lim_{\substack{x\rightarrow -2}} \frac {1 - x - 2x^2} {x + 1} = 5\lim_{\substack{x\rightarrow -2}} \frac {1 - x - 2x^2} {x + 1} = 5 if \forall \epsilon &gt; 0, \exists \delta &gt; 0 : 0 &lt; |x + 2| &lt; \delta \Rightarrow |-2x - 4| &lt; \epsilon

Investigation to find \delta
\begin{align*}<br /> |-2x - 4| &lt; \epsilon\\<br /> \Rightarrow |2x + 4| &lt; \epsilon\\<br /> \Rightarrow 2|x + 2| &lt; \epsilon\\<br /> \Rightarrow |x + 2| &lt; \frac {\epsilon} {2}<br /> \end{align*}

Therefore, choose \delta = \frac {\epsilon} {2}

Proof:
Let \epsilon &gt; 0 and set \delta = \frac {\epsilon} {2}

\begin{align*}<br /> |x + 2| &lt; \delta\\<br /> \Rightarrow |x + 2| &lt; \frac {\epsilon} {2}\\<br /> \Rightarrow 2|x + 2| &lt; \epsilon\\<br /> \Rightarrow |2x + 4| &lt; \epsilon\\<br /> \Rightarrow |-2x - 4| &lt; \epsilon<br /> \end{align*}
As required