April 8th, 2008, 04:28 AM  #1 
Newbie Joined: Apr 2008 Posts: 1 Thanks: 0  A limit I can't find ..
Hi there. I'm new to this forum because I had a question yesterday I couldn't figure out.. Here's the problem: 'Find the limit of x approaching 0 of (x^1/3)  (x^1/5)  x I tried lots of methods to figure this out. L'hopital is out of the question (because our professor at my university said so). I really have no idea and need help desperately .. Thanks in advance! Greets 
April 8th, 2008, 08:55 AM  #2 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Greetings: Factoring x^(1/5) from the numerator leaves, limit [(x^(1/5))*(x^(2/15)  1) / x] = limit [(x^(1/5)) / x] * limit (x^(2/15)  1). = limit (1 / x^(4/5)) * limit (x^(2/15)  1) As x>0, (1 / x^(4/5)) and (x^(2/15)  1) tend toward infinity and 1 respectively. Hence, limit (x>0) [((x^(1/3)  x^(1/5)) / x] does not exist. Regards, Rich B.[/color] rmath4u2@aol.com 
April 10th, 2008, 04:29 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1385 
As x → 0, 1/x^(4/5) → ∞ and x^(2/15) − 1 → −1. Hence, (x^(1/3) − x^(1/5))/x → −∞ as x → 0. 
April 11th, 2008, 03:18 PM  #4 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Correct you are. In my experience, a limit is a real number. Hence if an expression tends toward infinity, its limit is said not to exist. It is a matter of semantics; the terminology depends upon personal choice, and which author one opts to follow. In general, the statements "lim(x>a)[f(x)] = ±infinity" and 'lim(x>a)[f(x)] DNE" are synonymous (the latter, of course, ignoring the specific nature of unboundedness). Regards, Rich B.[/color] rmath4u22aol.com 
April 11th, 2008, 04:45 PM  #5  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Quote:
Correct me if I'm wrong, but: limit {as (x,y)>(0,0)} [(x^2y^2)(x^2+y^2)] Does not exist, but it does not converge to infinity. Along the x axis, f(x,y)>1; along the yaxis, f(x,y)>1, thus: no value, in general. Of course, my calculus could use some brushing up, and my analysis is nonexistent...  
April 12th, 2008, 04:43 AM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
cknapp is right: bounded functions that don't 'settle down' are divergent. (Although he should have used the phrase "diverge to infinity" rather than "converge to infinity".) 
April 16th, 2008, 10:40 AM  #7 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]This is true. My point, and I should have said so initially, is that a limit that goes to infinity is rightfully said not to exist. The converse is not necessarily true as can be seen by the simple example: Limit(x>inf)[sin(x)]. Clearly the limit does not exist; but the argument goes to neither infinity nor the negative thereof. Regards, Rich B.[/color] rmath4u2@aol.com 

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