Determine whether the sequence converges or diverges. If it converges, find the limit.
$ a_n = n \sin (1/n) $
converges to 1
Okay, so this is ahh type of problem where, when we're looking At Ltd's In Approaches Infinity of a. M. If we were to just plugin and equals infinity than we have Infinity times zero. That's indeterminate form, but it's an indeterminate form that if we just manipulate a little bit, then we can hope, maybe be able to use low Patel's rule. That's it. We'LL try and do here. So we'll figure out how we can rewrite it so that Low Patel's rule works. So instead of multiplying by and we can just divide bye one over in Dividing is the same thing as multiplying by the reciprocal. Okay, And now, if we just were toe plug in and equals Infinity than we'd have zero over zero. Which is, of course, not allowed. But if you are in a setup like that, then you are allowed to use local house rule says it will do here. It's after we do. Local towers rule solo Patel's rule. Remember, we do the derivative on top, divided by the derivative in the denominator. So the derivative of sign of one over end with with respect to end. So that's going to be co sign of one over in and then chain rule. So we take the derivative of one over in, So that's minus into the minus two. And then we're dividing by the derivative of one over end, which, as we commentate, is minus in to the power of minus two. So these guys will cancel and we get limit as n goes to infinity of co sign of one over end and again Cosenza continuous functions. You can pull the limit inside and get co sign of limit as n goes to infinity of one over in co sign of zero, we get one. So we do converge and we converge to one hands to the main trick that we did here. Wass noticing that if we were to plug in and equals, Infinity would get infinity Times zero, and that something that is in that form can be re written in such a way to wear low. Patel's rule makes sense. Remember, Low Patel's rule works with fractions. If you have fractions where plugging in the limit gives you infinity over infinity or zero over zero, then you could do the trick where you do derivative on top, divided by the derivative on the bottom. That's what we did here