Solve L'Hopital's Rule for y(t) at Limit of t to Infinity

[dm164]

So I've come across this formula that I derived. y(t) =v²t/√(v²t²+b²)

I would like to solve the limit of t to infinity analytically. When I apply L'Hopital I get
y = lim v² / lim v²t/√(v²t²+b²)

but as you can see I would have to apply L'Hopital rule an infinite amount of times, now I don't know if you say it becomes x/(x/(x/..))). with x= v² whatever value that is.

By inspection of a grapher I would say it's v , it also looks like v*sin(arctan(v/b*t)), which at t-> arctan -> pi/2 then sin() -> 1 so the answer is v. But, how do I know arctan(t) t->inf it pi/2 besides geometrically it makes sense.

Any ideas about the infinite L'Hopital, or infinite divisions how something like that could be solved.

 

[Simon Bridge]

You mean find $L:$ $$L=\lim_{t \rightarrow \infty}\frac{v^2t}{\sqrt{v^2 t^2 +b}}$$

I'd be inclined to take the vt out from under the square root. $b/v^2t^2\rightarrow 0$.

Under L'Hopital, if you just keep repeating it, I suspect you'll and up finding L=f(L) ...and solve for L.

 

[CompuChip]

I would like to solve the limit of t to infinity analytically. When I apply L'Hopital I get
y = lim v² / lim v²t/√(v²t²+b²)

You mean that you get
\lim_{t \to \infty} y = \frac{\lim_{t \to \infty} v^2}{ \lim_{t \to \infty} \left( \frac{v^2 t}{\sqrt{v^2 t^2 + b^2}} \right) }
(note I put the missing limit on the left hand side), i.e.
\lim_{t \to \infty} y = \frac{\lim_{t \to \infty} v^2}{\lim_{t \to \infty} y} ?

In Simon's notation, that would be
L = \frac{\ldots}{L}
except that you don't have to apply L'Hopital more than once.

 

[Simon Bridge]

The same trick is often used in integration by parts.
Y'know, I'd never heard of L'Hopital before I started at this forum. It would have been handy. But in this case, do you really need it? Oh well, never mind.

Aside:

The only stupid question is the one you don't ask. The only stupid person is the one who pretends to know.

"Remember children: there are no stupid questions, only stupid people."
-- Mr Garrett (Southpark)​

 

[dm164]

I'd be inclined to take the vt out from under the square root. $b/v^2t^2\rightarrow 0$.

Ah yeah, didn't see that way.

You mean that you get

\lim_{t \to \infty} y = \frac{\lim_{t \to \infty} v^2}{\lim_{t \to \infty} y} ?

In Simon's notation, that would be
∴L = \frac{\ldots}{L}
except that you don't have to apply L'Hopital more than once.

Oh sure, kind of seems silly now \lim_{t \to \infty} y =√(\lim_{t \to \infty} v^2)= v

Always good to know a fews ways to an answer.