<p>Please explain to me why lim x-> infinity (x*sin(1/x))=1.
When I tried to do it, I split it into lim x->inf x * lim x->inf sin(1/x).
The first side equals inf, and the second equals 0. It seems that the answer should be 0 because of the multiplication. But the infinity involved makes that wrong, and the answer 1, but I'm not sure how you get there sans calculator.</p>
<p>Oh this is a classic!
Here is what you do:
lim x-> infinity (x*sin(1/x))=
lim x-> infinity (sin(1/x) ) / (x^-1)
Then use L'hopital's Rule:
lim x-> infinity (-x^-2 * cos(1/x) ) /( -x^-2)
-x^-2 cancell and lim x-> infinity cos(1/x) = 1</p>
<p>or even faster than that lim x -> infinity (x*sin(1/x)) = lim x -> infinity (sin(1/x)) / (1/x). Let u = 1/x, then the limit is transformed to lim u -> 0 sin u / u = 1.</p>
<p>Thanks, and I get it, but the problem is, this is in a packet we got to work on before we learned derivatives, and we haven't actually learned l'Hopital's rule yet. (I just looked it up.)</p>
<p>Edit: Thanks stupak for the other way.</p>
<p>NO! Stupak, your method is wrong, don't show this guy a wrong method. 0/0 is not 1!!!!!!!!! In fact, if people knew what 0/0 was, then our lives would be much easier. Again saying 0/0 = 1 is like saying 0*infinity = 0 or 1^infinity = 1. Then only way to solve it, probably, is trough L'Hopital's.</p>
<p>Well first of all, I'm a girl. Second, Stupak never said 0/0=1. Stupak simply used substitution in a way that was very similar to another problem I did in class. And there has to be a way besides l'hopital's because it was in a packet of questions specifically designed for an early chapter that comes before any kind of derivative.</p>
<p>Sorry about calling you a guy, that was unfounded. I see what it is:
lim u -> 0 sin u / u = 1 technically that is sin(0)/(0) = 1
Yes, I see that probably you are supposed to used knowledge that lim x--> 0 sin (x)/x = 1. They introduce the fact yearly on, but the only way to proove this is by L'Hopital's. Yeah, you are right I guess you can use that fact, just understand that you dont really know what lim x -> 0 sin (x) /x is untill you get to L'hopital's rule.</p>
<p>Yeah, you usually don't learn L'hopital's rule before or during limits.</p>
<p>Actually, you can prove that lim x->0 sinx/x = 1 with the Squeeze Theorem, which basically "squeezes" the graph of sin(x)/x between a graph that is always less than sinx/x and one that is always greater (at least near 0, anyway).
The one thing that bothers me about stupak's method is that we're trying to find a limit at infinity. Why, then, do we use lim x-->0 to find the limit when the limit is at infinity, not 0?</p>
<p>Recall my substutition u = 1/x. In order to translate the limit into the world of u, you have to replace x --> infinity with u --> something, which also means 1/x --> something. 1/x approaches 0 as x --> infinity, thus u --> 0 as x--> infinity</p>
<p>he used the substitusion, that's why its zero. But the Squeeze Theorem is pretty lame, I never liked it, at least not as much as L'hopital's.</p>
<p>well, l'hopital's is a pretty complex concept, and this is beginning single-variable calculus. don't get me wrong, using l'hopital's rule in practice is easy, but understanding why it works always isn't that easy.</p>
<p>oh yeah, I agree, in fact I don't even remember how to prove the theorem at this point, but it's a hell of a lot of fun to use.</p>