Question

Epsilon delta proof help. Problem in comments

Answer 1

try \(\delta = \epsilon\)

Answer 2

I don't even know what that means. The video my teacher made me watch made no sense. I looked for more but just got more confused.

Answer 3

@katiestyles212, I'll do the bulk of the work but I'll leave you to write out all the frilly bits. If you have a basic understanding of limits, my working should make sense to you:

\[\left| \frac{ 9-4x^2 }{ 3+2x } -6\right|\]
\[\left| \frac{ 9-4x^2 }{ 3+2x } -\frac{6(3+2x)}{(3+2x)}\right|\]
\[\left| \frac{ 9-4x^2 -6(3+2x)}{(3+2x)}\right|\]
\[\left| \frac{ -4x^2-12x-9}{(3+2x)}\right|\]
\[\frac{ 4x^2+12x+9}{(3+2x)}\]
\[\frac{ (2x+3)^2}{(3+2x)}\]
\[2x+3\]

At x=1.5, this equals 6 (the limit).

I know this is far from a delta epsilon proof, but I've done the hard bit and hopefully you can do the rest :)

Answer 4

say \(f(x) = \frac{9-4x^2}{3+2x}\)

you want to show for each \(\epsilon \gt 0\), there always exists a \(\delta \gt 0\) such that :

\[|x-(-1.5)| \lt \delta \implies |f(x)-L| \lt \epsilon\]

Answer 5

lol tom, that approach seemed like a bit of extra work :3
no need to find a common denominator.
factor your difference of squares and cancel out with the denominator.
just another way to get the same result c:

Answer 6

I know @zepdrix, but a lot of these limits require a common denominator so it's good to know :) I should've mentioned the more direct approach though, so thanks for posting it!

Answer 7

if you have managed to go through tom's work, you would see that \(\large \delta = \epsilon /2\) will also work because :

\[0 \lt |2x+3| \lt \epsilon \]

\[0 \lt |x+1.5| \lt \epsilon/2 \]

\[0 \lt |x-(-1.5)| \lt \epsilon/2 \]

Answer 8

Notice that middle expression is exactly what you're looking for : the distance from -1.5 to any x :

if you let \(\delta = \epsilon / 2\), you're done.

Answer 9

actually it turns out that \(\delta = \epsilon\) wont work, you must take to be less than \(\epsilon /2 \)