Question

can someone tell me how to prove for infinite limit with epsilon and delta or N and M?

Answer 1

Yes, if you gave us an example, we could.

Answer 2

What's the function?

Answer 3

l\[\lim_{x \rightarrow \infty}3x^{2}-x-2/5x^{2}+4x+1\]

Answer 4

\[\lim_{x \rightarrow \infty}~\dfrac{3x^{2}-x-2}{5x^{2}+4x+1}\]

like that ?

Answer 5

Yes, like that

Answer 6

so what does it evaluate to ?

Answer 7

you need to know the limit before starting to prove it using epsilon-delta

Answer 8

i know the limit it's 3/5

Answer 9

but the prove is quite confusing.

Answer 10

its easy once you know how to play the game

Answer 11

i guess... so :)

Answer 12

Let \(f(x) =\dfrac{3x^{2}-x-2}{5x^{2}+4x+1} \)

given any positive number \(\epsilon\), you have to show that there exists some \(x\) value, after which the difference between \(f(x)\) and \(\dfrac{3}{5}\) is less than \(\epsilon\).

Answer 13

Given any positive number \(\epsilon\), prove that there exists some \(x\) value beyond which below is true :
\(\left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\lt \epsilon\)

Answer 14

simplify and try expressing \(x\) in terms of \(\epsilon\)

Answer 15

@kal1921

Answer 16

simplify? sorry where is delta?

Answer 17

Simplify so that you can find the bound on delta. You have to work backwards

Answer 18

yeah what we're doing is just scratch work to express \(x\) in terms of \(\epsilon\)

Answer 19

\[-\epsilon< equation<\epsilon\]
i get this...what is next?

Answer 20

for \(x\gt 1\), we have :

\(
\left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\\~\\
=\left|\dfrac{17x+13}{5(5x^2+4x+1)}\right|\\~\\
\lt\left|\dfrac{17x+13}{25x^2}\right|\\~\\
\lt\left|\dfrac{17x+17x}{25x^2}\right|\\~\\
=\left|\dfrac{34}{25x}\right|\\~\\
\)

\(\dfrac{34}{25x} \lt \epsilon \implies x \gt \dfrac{34}{25\epsilon}\)

therefore \(x = \dfrac{34}{25\epsilon}\) works, now you can start the proof.

Answer 21

sorry, how can i relate x and delta... if it x wasn't approaching to infinity i would say
/x-a/>delta

Answer 22

The definition for limits at infinity is slightly different :
\(\lim\limits_{x\to\infty}f(x)=L\) means that for "every" \(\epsilon\gt 0\), there exists some \(x\) value, \(N\), such that \(|f(x)-L|\lt\epsilon\) for all \(x\gt N\).

Answer 23

In present problem, we have cooked up that \(N\) to be \(\dfrac{34}{25\epsilon}\)

Answer 24

you choose "any" positive \(\epsilon\),

the value \(|f(x)-\frac{3}{5}|\) will always be less than that chosen \(\epsilon\) whenever \(x\) is greater than \(\dfrac{34}{25\epsilon}\)

Answer 25

In other words, we can make \(f(x)\) as close to \(\frac{3}{5}\) as we want by making \(x\) sufficiently large.

Answer 26

I think this is same as the N and M definition that you were refering to in main question