Hi Veeky Forums...

Hi Veeky Forums, we have been studying in my class the epsilon delta definition of limits and I have done a couple exercises, but I haven't found a way to solve the one in the pic. Could any mathfags in here help me out please?

I would try L'Hôpital's rule.

After two applications, the numerator would contain a constant and the denominator would contain a term with t. Thus as t goes to infinity, [math]a_t[/math] will go to zero

I know that I could get the result through L'Hôpital's rule, but our teacher specifically asked us to do the proof through epsilon delta definition. Do you have any idea how to do it that way? Thanks for your help anyway.

I think he's talking about the formal definition of limit.

I've just finished analysis 1 OP, but it's been a while since I've done this (I only did that in calc 1). If it's not trigonometric, then it shouldn't be hard as far as I remember.

Good luck.

YES, that's exactly what I'm looking for. The problem that I'm having is that the variable tends to infinity, and even after doing the change of variable I can't see how to solve for delta in terms of epsilon.

Fuck of with your l'Hôpital's rule. That shit meme is far from necessary 99% of the time.

Just develop the denominator, divide on both sides by [math]t^3[/math] and apply the basic theorems on operations between limits.

It is as simple as setting [math]a_t[/math] equal to epsilon and solving for the corresponding delta.

I have tried to solve it that way or through partial fractions decomposition, but I still can't solve for delta.

Except solving third degree polynomials is much harder than second degree ones (which I assume is OP's level on polynomials).

Do you want the optimal delta? If so, solving it as a polynomial might be the only way, but as I've just said, it's tedious. If you don't care about having everything optimal, my method works for sure although it would need some careful planning.

Bound it by suitable constant multiplied by 1/t.

Yes your right, finding the optimal delta would be too complicated and I don't want to solve a third degree polynomial. What do you mean by careful planning?

[eqn]
t^2 +4 = t^2 + 4t + 4 - 4t = (t + 2)(t + 2) - 4t
[/eqn]
so you have
[eqn]
a_t = \frac{t+2}{(t-1)(t-3)} - \frac{4t}{(t-1)(t+2)(t-3)}
[/eqn]
note that [math]t+2 = t + 3 - 1[/math] and [math]4t = 4t -4 + 4[/math] so
[eqn]
a_t = \frac{1}{t-3} + \frac{3}{(t-1)(t-3)} - \frac{4}{(t+2)(t-3)} - \frac{4}{(t-1)(t+2)(t-3)}
[/eqn]
the limit is clear, this was literally just fraction manipulation
t. mathematician

This requires the reader to have proven that the limit of sums is the sum of limits.

that's super easy for the guy to do on his own as a lemma for the problem

what's that? you want me to show you just how easy that is?

[math]\textbf{Lemma.}[/math] If [math]\lim_{t \to \infty} f(t) = a[/math] and [math]\lim_{t \to \infty} g(t) = b[/math] then [math]\lim_{t \to \infty} f(t) + g(t) = a + b[/math].

[math]\textit{Proof:}[/math] Let [math]\epsilon > 0[/math]. Then [math]\exists t_f, t_g[/math] such that [math]\forall t > t_f, \left|f(t) - a\right| < \frac{\epsilon}{2}[/math] and [math]\forall t > t_g, \left|g(t) - b\right| < \frac{\epsilon}{2}[/math]. Let [math]t_M = \max\{t_f,t_g\}[/math]. Then [math]\forall t > t_M[/math] we have [math]\left|f(t) + g(t) - a - b\right| \leq \left|f(t) - a\right| + \left|g(t) - b\right| < \epsilon[/math].

t. mathematician

[eqn]a_t = \frac{(t-2)(t+2)}{(t-1)(t+2)(t-3)}[/eqn]
[eqn]a_t = \frac{(t-2)}{(t-1)(t-3)}[/eqn]
[eqn]a_t = \frac{(t-2)}{(t-2)(t-2) - 1}[/eqn]
[eqn]a_t = \frac{1}{(t-2) - \frac{1}{(t-2)}}[/eqn]
[eqn]\lim_{t \to \infty} \frac{1}{(t-2)} = 0[/eqn]
[eqn]\lim_{t \to \infty} a_t = \frac{1}{(t-2) - 0}[/eqn]
[eqn]\lim_{t \to \infty} a_t = \frac{1}{(t-2)}[/eqn]
[eqn]\lim_{t \to \infty} = 0[/eqn]

what are the roots of t^2 + 4 user?

Oops.
Fixed.

[eqn]a_t = \frac{(t-2)(t+2)+8}{(t-1)(t+2)(t-3)}[/eqn]

[eqn]a_t = \frac{(t-2)+\frac{8}{(t+2)}}{(t-1)(t-3)}[/eqn]

[eqn]a_t = \frac{1+\frac{8}{(t+2)(t-2)}}{(t-2)-\frac{1}{(t-2)}}[/eqn]

[eqn]a_t = \frac{1+\frac{8}{t^2-4}}{(t-2)-\frac{1}{(t-2)}}[/eqn]

[eqn]\lim_{t \to \infty} \frac{1}{(t-2)} = 0[/eqn]

[eqn]\lim_{t \to \infty} \frac{8}{t^2-4} = 0[/eqn]

[eqn]\lim_{t \to \infty} a_t = \frac{1+0}{(t-2)-0}[/eqn]

[eqn]\lim_{t \to \infty} a_t = \frac{1}{(t-2)}[/eqn]

[eqn]\lim_{t \to \infty} a_t = 0[/eqn]

I think the key to these is to do rough estimates that simplify the expression for example
[math]a_t

he is no closer to proving the limit with your post, your post is unrelate

What? To prove that a_t-> 0 when t-> infinity with epsilon delta you have to show that for every d>0 there exists a N>0 such that |a_t-0|=a_tN, this is exactly what I've done, well, my implications should be bi-implications but that's the only issue. Also, this is the only attempt at a direct epsilon delta in the thread it seems. So even if it's faulty, it's currently the most relevant.

the only estimates you need are these:
for large t,
t^2 < t^2 + 4 < 2t^2
and
t^3 > (t-1)(t+2)(t-3) = (t-1)(t^2-t-6) = t^3-2t^2-5t+6 > t^3-7t^2
so 1/t ≤ a_t < 2/(t-7) which converges to zero on both sides