- #1
dm164
- 21
- 1
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.
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.
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