Limit as x tends to infinity

I need to prove that 2x/x-1 tends to 2 as x tends to infinity using the epsilon delta method. I don't understand how you select K. The answer is below if anyone could explain it that would be great. :smile:

Reply 1

LelouchViRuge

7

Original post by Lunu

I need to prove that 2x/x-1 tends to 2 as x tends to infinity using the epsilon delta method. I don't understand how you select K. The answer is below if anyone could explain it that would be great. :smile:

I hate this -_-

try watching this

Reply 2

Zacken

22

Original post by Lunu

I need to prove that 2x/x-1 tends to 2 as x tends to infinity using the epsilon delta method. I don't understand how you select K. The answer is below if anyone could explain it that would be great. :smile:

Which bit in particular don't you understand? There isn't really a need to choose a specific value of

k

as long as we pick one that satisfies the fact that when

x>k

then

|f(x) - \ell| < \epsilon

.

Sometimes, if you can't spot a useful value of k, then work backwards.

i.e: assume that

Unparseable latex formula:

\displaystyle[br]\begin{align*} \frac{2}{x-1} < \epsilon &\Rightarrow 2 < (x-1)\epsilon\\ & \Rightarrow \frac{2}{\epsilon} < x-1 \\ & \Rightarrow x > 1 + \frac{2}{\epsilon}\end{align*}

.

So now you can scrawl this working off and start the proof properly with "choose

k > 1 + \frac{2}{\epsilon} \Rightarrow \cdots

" because you've found the 'necessary' value of

k

in your 'working backwards' thingy which is useful for computation but proves nothing by itself.

Reply 3

Lunu

OP

1

Original post by Zacken

Which bit in particular don't you understand? There isn't really a need to choose a specific value of

k

as long as we pick one that satisfies the fact that when

x>k

then

|f(x) - \ell| < \epsilon

.

Sometimes, if you can't spot a useful value of k, then work backwards.

i.e: assume that

Unparseable latex formula:

\displaystyle[br]\begin{align*} \frac{2}{x-1} < \epsilon &\Rightarrow 2 < (x-1)\epsilon\\ & \Rightarrow \frac{2}{\epsilon} < x-1 \\ & \Rightarrow x > 1 + \frac{2}{\epsilon}\end{align*}

.

So now you can scrawl this working off and start the proof properly with "choose

k > 1 + \frac{2}{\epsilon} \Rightarrow \cdots

" because you've found the 'necessary' value of

k

in your 'working backwards' thingy which is useful for computation but proves nothing by itself.

Thank you so much. It seems so simple now! You have no idea how much you've helped! :smile:

Reply 4

Zacken

22

Original post by Lunu

Thank you so much. It seems so simple now! You have no idea how much you've helped! :smile:

No worries, it's also really easy once you work backwards because then for the actual proof, you can just reverse everything you did in the scrap work and it's immediate. :biggrin:

Reply 5

Lunu

OP

1

Original post by Zacken

No worries, it's also really easy once you work backwards because then for the actual proof, you can just reverse everything you did in the scrap work and it's immediate. :biggrin:

Sorry for asking another question but if you were asked to find the negation of the limit for x tends to infinity to prove that the limit does not exist. How do you choose epsilon?

Reply 6

Zacken

22

Original post by Lunu

Sorry for asking another question but if you were asked to find the negation of the limit for x tends to infinity to prove that the limit does not exist. How do you choose epsilon?

Well - if you negate the limit definition you get (in words)

There exists a positive real

\epsilon

such that for all delta there exists an

x\in \mathbb{R}

such that

0 < |x-a|< \delta

implies

|f(x) -\ell | \geq \epsilon

So - given this, can you try working backwards for yourself and see what you get?

From what I've seen, you usually just pick

\epsilon = 1

and show that

|f(x) - \ell| > 1

by picking a suitable

\delta

.

(edited 7 years ago)

Limit as x tends to infinity

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