Hi Veeky Forums...

Hi Veeky Forums, so about three days ago I made a thread asking people to help me with a problem related with an epsilon-delta proof and a guy posted a way to do it. However, I made the mistake of thinking that the thread would still be there after I came back from a class. I remember how he did it for the most part, but I forgot the end of the proof and I've got to a point where I'm stuck and I don't know what to do next. Could anyone help me? Or is the guy that helped me last time still there?

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dude this is so wrong

keep in mind that you want to show that the expression is similar to t^2 / t^3 or 1 / t for sufficiently large t
dont be afraid of using words to explain what you are doing in your working

no fucking idea what happened after |(t+2)/((t-1)(t-3))| desu

Do you need to use a direct epsilon delta argument or are you able to use theorems? Because with just the squeeze theorem you can do this without much effort and just conclude with an epsilon delta argument.

This isn't actually very hard right? I'm a physicist so my way of thinking may not be absolutely rigorous:
working out the brackets yields a term of [math] t^3 [math/]. If you then divide above and below the denominator by [math] t^2 [math/] the term above will go to 1, and the term below is simply t, as the [math] \frac{t^3}{t^2} [math/] is t.
now letting t appoach infinity, all our terms with 1/t go to 0, as will our function, since that too is simply 1/t.

Why the fuck do you want to prove that with epsilon
Just solve it like everyone
[math]a_t=\frac{t^2+4}{(t-1)(t+2)(t-3)}\sim\frac{1}{t}\underset{t\to +\infty}{\longrightarrow}0[/math]
And be done with it

You can't do that with a delta epsilon proof.

You have to show that the terms in the sequence eventually tend to some number. You have to start the proof by the logical quantifier. If the first line of this proof isn't
"Assume epsilon greater than 0" it's automatically wrong.

The next and hardest part is finding a tail of the sequence so that you know the sequence converges. It's not clear to me what this is, but you have to start by removing the absolute value bars, which will result in an inequality that you use for N. From there, the proof is straightforward plug and chug.

I think i've got it:
We have the sequence: [math]a_{t} : \mathbb{N} \setminus {1,3} \longrightarrow \mathbb{R} [/math] .
We want to show that: [math] \forall \epsilon > 0 : \exists N(\epsilon) > 0 : \forall t>N(\epsilon) : |a_{t}|< \epsilon [/math] .

Now,
[math] \left|\frac{t^2+4}{(t-1)(t+2)(t-3)}\right| N_{1}(\epsilon)=3+ \lfloor \frac{1}{2 \epsilon} \rfloor \frac{5}{2}[/math]
[math] \Rightarrow t-3>\lfloor \frac{1}{2 \epsilon} \rfloor \frac{5}{2} [/math]
[math] \Rightarrow \frac{2}{5} (t-3) > \frac{1}{2 \epsilon} [/math]
[math] \Rightarrow \frac{5}{2} \left| \frac{1}{t-3} \right|N_{2} (\epsilon)=1 + \lfloor \frac{1}{ \epsilon} \rfloor \frac{3}{2}[/math]
[math] \Rightarrow \frac{2}{3}(t-1)>\frac{1}{ \epsilon}[/math]
[math] \Rightarrow \frac{3}{2} \left| \frac{1}{t-1} \right|< \epsilon [/math]

Now, let [math] N( \epsilon) := min{N_{1}( \epsilon),N_{2}( \epsilon)} [/math]

then, [math]\forall t>N( \epsilon) [/math]:

[math]\left| a_{t} \right|

(this is my prof from calc 1 and 2 in the picture)