Question

Use epsilon-delta to prove
\[lim_{x\rightarrow -1}\dfrac{x+5}{2x+3}=4\]
Please, help

Answer 1

\[
|x-1|<\delta \implies \left|\frac{x+5}{2x+3}-4\right|<\epsilon
\]

Answer 2

You start here.

Answer 3

no, |x+1|, not x-1

Answer 4

Ok.

Answer 5

\[
|x-(-1)|<\delta \implies \left|\frac{x+5}{2x+3}-4\right|<\epsilon
\]

Answer 6

yes, but I got stuck at argue \(|\dfrac{-7}{2x+3}| \)

Answer 7

Can you simplify the inequalities a bit?

Answer 8

yes, it is \(|\dfrac{-7}{2x+3}||x+1|<\varepsilon\)

Answer 9

Okay now then, You want to make sure that \[
\left|\frac{-7}{2x+3}\right|\delta <\epsilon
\]Right?

Answer 10

First thing you want to do is figure out what values the function will be when \(x\) is near \(-1\).

Answer 11

it is 4

Answer 12

Alternatively, if there is a maximum value, you want to find the maximum value of it.

Answer 13

\(|x+1|<\delta \\-\delta< x+1<\delta\\-1-\delta<x<-1+\delta\)
If \(\delta =1,\) then -2< x< 0
hence -4< 2x < 0
and -1< 2x +3 <3
That gives me \(\dfrac{1}{2x+3}< -1\) or \(\dfrac{1}{2x+3}>3\)

Answer 14

and how to put -7 in to those intervals, and take absolute value to get the expression < epsilon?

Answer 15

Actually, you should look at this first:
http://www.wolframalpha.com/input/?i=%7C-7%2F(2x%2B3)%7C

Answer 16

I think a good minimum range for delta would be something like \(1/4\). If you get to \(-1.5\) then we get into trouble.

Answer 17

So in the range: \[
-1/4 <x+1<1/4
\]What sort of range do we see for \[
\left|-\frac{7}{2x+3}\right|
\]

Answer 18

When \(\delta = 1/4\), when we see that max value of it will be \(14\).

Answer 19

Is there any logic way to have delta or just guess? Since we are not allowed to use math tool in class, how can I get that value?

Answer 20

We can say \(\delta = \min(1/4, f(\epsilon))\) so that \(\delta\) neger gets to large.

Answer 21

What if 1/4 is too small?

Answer 22

Now, if \(x\) gets really close to \(-1\), we know that \[
\left|\frac{-7}{2x+3}\right|<14
\]

Answer 23

We have this relationship: \[
\delta <\frac{\epsilon }{\left|\frac{-7}{2x+3}\right|}
\]

Answer 24

\[
\delta <\frac{\epsilon}{14}\leq \frac{\epsilon }{\left|\frac{-7}{2x+3}\right|}
\]

Answer 25

So we can safely say \[
\delta = \min (1/4, \epsilon/14)
\]

Answer 26

then, take delta = min {1/4, epsilon/14} and go backward, right?

Answer 27

I got it. Thanks a lot. :)

Answer 28

I suppose so. That would be a big rigorous.

Answer 29

lost connection again :(