Another calculus quandry.

Hsu · 10-24-2007, 02:52 AM

This one's about to make me crazy. I don't know what I'm missing.

It's the limit as x approaches infinite of x*tan(1/x). Graphing it shows that it should approach the same thing as tan(1), and obviously getting tan(x/x)=tan(1). But I can't for the life of me seem to find any property that lets me just go to that step, and expanding the equation into sines and cosines just ends up with the whole thing coming out to 0. Helps, please.

Hsu · 10-24-2007, 04:12 PM

Nevermiiiind.

Yahoo answers got it for me.

Pureghetto · 10-24-2007, 06:45 PM

mind poasting the 'proof' so I don't have to ask my friend and have him go 'sigh I taught you this a year ago' ? :D

Weimar Pluto Knight VII · 04:29 AM

lim (x*tan(1/x)) = lim (tan(1/x) / (1/x)) by L'Hospital's Rule = lim ((sec (1/x))^2) * (1/x^2)/(1/x^2) = lim (sec(1/x))^2 = 1.

Hsu Cid's Knight Zell 007 Location: Ohio	#2 (permalink) Nevermiiiind. Yahoo answers got it for me.
10-24-2007, 04:12 PM

Pureghetto Itadakimasu~ Location: Barrie, ON	#3 (permalink) mind poasting the 'proof' so I don't have to ask my friend and have him go 'sigh I taught you this a year ago' ? :D
10-24-2007, 06:45 PM

Weimar Pluto Knight VII Location: Michigan, USA	#4 (permalink) lim (xtan(1/x)) = lim (tan(1/x) / (1/x)) by L'Hospital's Rule = lim ((sec (1/x))^2) (1/x^2)/(1/x^2) = lim (sec(1/x))^2 = 1. Last edited by Weimar Pluto Knight VII; 01-20-2008 at 04:29 AM.
11-16-2007, 07:26 PM