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April 8th, 2008, 04:28 AM   #1
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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)

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!
Katano is offline  
April 8th, 2008, 08:55 AM   #2
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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.


Rich B.[/color]
nikkor180 is offline  
April 10th, 2008, 04:29 AM   #3
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As x → 0, 1/x^(4/5) → ∞ and x^(2/15) − 1 → −1.
Hence, (x^(1/3) − x^(1/5))/x → −∞ as x → 0.
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April 11th, 2008, 03:18 PM   #4
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[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).


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April 11th, 2008, 04:45 PM   #5
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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...
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April 12th, 2008, 04:43 AM   #6
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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".)
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April 16th, 2008, 10:40 AM   #7
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[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.


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