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 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
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Quote:
 Originally Posted by nikkor180 In general, the statements "lim(x-->a)[f(x)] = ±infinity" and 'lim(x-->a)[f(x)] DNE" are synonymous
While I agree with your post in general, this is not necessarily true, unless I"m mistaken.

Correct me if I'm wrong, but:
limit {as (x,y)->(0,0)} [(x^2-y^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 y-axis, f(x,y)->-1, thus: no value, in general.

Of course, my calculus could use some brushing up, and my analysis is non-existent...

 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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