Limits Approaching InfinityDate: 02/16/2002 at 12:34:58 From: Miranda Cir Subject: Limits approaching infinity I am lost when it comes to limits approaching infinity. Is the only way to answer them by memorizing all the general questions you may come across? ex: lim --> infinity (1/2)^n = 0 and lim --> infinity 4^(-infinity) = infinity and lim --> infinity 4^[1/(-infin)] = 1 I'm confused. Can you give me some advice? Date: 02/16/2002 at 23:09:00 From: Doctor Peterson Subject: Re: Limits approaching infinity Hi, Miranda. First, we have to correct what you wrote. A limit has to state what variable is going to infinity. In the first, it is presumably n; but the others have no variable. I think you meant lim[n->inf] (1/2)^n = 0 lim[n->inf] 4^-n = inf lim[n->inf] 4^(1/-n) = 1 There are only a couple of basic facts you should learn; one is: lim[n->inf] 1/n = 0 This says that if you divide by larger and larger numbers, the quotient goes to zero. Memorize the shape of the graph of this reciprocal function; it's extremely useful and common. Let's take your examples one at a time. lim[n->inf] (1/2)^n = lim[n->inf] 1/(2^n) = lim[m->inf] 1/m = 0 I first simplified the exponent using exponent properties so I had the reciprocal of something. Then I noticed that 2^n goes to infinity when n goes to infinity; by calling that m, I have the limit of 1/m, which is 0 according to the fact above. A general fact worth knowing is that a^n goes to zero if a<1, and to infinity if a>1. You might want to graph this for several values of a to see what it means. Memorize the shape of the graph in these two cases. (And see what happens if a=1, too.) Your second problem: lim[n->inf] 4^-n = lim[n->inf] 1/(4^n) = 0 by the same reasoning; it's just written differently at first. Did I correct the problem correctly? Finally, lim[n->inf] 4^(1/-n) = lim[n->inf] 4^(-1/n) = lim[p->0]4^-p = 4^-0 = 1 This time I looked at the exponent -1/n and saw that it goes to zero; since 4^0 is 1, we don't need any tricks to find the limit. It looks as if a general principle is: look at the base and the exponent, and see where they are going. As I said before, if the exponent is going to infinity, the limit depends on the base. But if the exponent is going to zero, the limit is 1 unless the base is, or is going to, 0, which would be indeterminate. If you need more help, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 02/17/2002 at 12:33:25 From: Miranda Subject: Limits approaching infinity I just wanted to thank you! This website is such an amazing help, I am definitely going to be coming back for more questions! Thanks again. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/