Calculus challenge...

[AtmaWeapon]

I have a rather odd request.

I took an exam in calculus today. I encountered the problem:



lim 5x*sin(3/x)
x->infinity


It seemed simple at first; 3/x will approach 0 as x approaches infinity. Therefore, sin(3/x) would approach 0 as x approached infinity. Since sin(3/x) evaluated to 0, I decided the correct answer was 0.

HOWEVER, things were not so simple. I plugged the equation into my TI-89, and somehow the calculator got 15 for the answer.

So, who's right, me or the calculator? If it's the calculator, please post an explanation.

[matthew986]

Well I dont know with my mad Geometry skilz. But the way you explained it you would be corect.

NOTE: I DONT KNOW WHAT IM TALKING ABOUT

[slothman]

The problem is that you are doing a limit of:
5*infinity*sin(3/infinity)
which equals (somewhat):
infinity * sin(3*0) = infinity * 0
which the limit doesn't like.
You have to use L'hopital's rule and derive both multipliers. Eventually you will get
5* lim 3*cos(3x)
x->0
which does indeed equal 15.

[Beldaran]

Originally posted by slothman
The problem is that you are doing a limit of:
5*infinity*sin(3/infinity)
which equals (somewhat):
infinity * sin(3*0) = infinity * 0
which the limit doesn't like.
You have to use L'hopital's rule and derive both multipliers. Eventually you will get
5* lim 3*cos(3x)
x->0
which does indeed equal 15.

And that concludes this weeks edition of, "Crap that I don't Understand!". Tune in next time when we discuss relationships.

[stormwatcheagle]

WTF! when was infinity a number! oh well, ill have to wait untill i go to calclus, i guess...

[AtmaWeapon]

WTF is L'Hopital's rule? I remember our professor going over it, but I thought he said we wouldn't need it for the exam. I could have sworn. Curse his Turkish accent.