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I know here we have the indeterminate form 0 * -∞ and therefor it must be changed to a quotient of the form 0/0 or ∞/∞ to apply l'hospital's rule.

So, I brought the lnx down to satisfy this condition:

lim x->0+ √x / (lnx)^-1 this shows the 0/0 form correct?

If so, then I can derive the top and the bottom to see if the result is workable.

lim x->0 (1/2)x^(-1/2) / -(lnx)^-2(1/x)

Am I on the right track here? I seems to get stuck around this point. Help!

2007-04-03 11:15:49 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

Thanks for the help everyone. Looking at this graphically 0 does not seem to be the limit. The book key says that the limit is 3, but again the graph does not appear to show this. Are we sure 0 is the limit here?

2007-04-03 12:31:19 · update #1

Nevermind about 3 being the book answer... I was reading off of the wrong section. The correct answer is indeed 0.

2007-04-03 13:07:13 · update #2

4 answers

lim ( √x ln(x) )
x -> 0+

This is of the form [0 * infinity], so it is an indeterminate form.
Instead of bringing the ln(x) down, how about you bring the
√(x) down instead? After all, √(x) = x^(1/2), and bringing it down would make it x^(-1/2).

lim ( x^(1/2) ln(x) )
x -> 0+

lim ( ln(x) / (x^(-1/2) )
x -> 0+

Now this is of the form [infinity/infinity], so use L'Hospital's rule.

lim ( (1/x) / [(-1/2)x^(-3/2)])
x -> 0+

First, move the x^(-3/2) up on top,

lim ( [(1/x)(x^(3/2)] / [(-1/2)] )
x -> 0+

We can now simplify the top, since (1/x) = x^(-1).

lim ( [x^(1/2) / (-1/2) ] )
x -> 0+

As x approaches 0 from the right, x^(1/2) approaches 0, so we can calculate the limit now.

0^(1/2) / (-1/2)

0/(-1/2)

0

Side note: The reason why it was better to bring down the square root of x instead of ln(x) is because the derivative of [ln(x)]^(-1) is more complicated. By leaving ln(x) in the numerator, at least when applying L'Hospital's rule it becomes simpler.

2007-04-03 11:22:53 · answer #1 · answered by Puggy 7 · 0 0

i'm assuming it is the right-exceeded reduce because the left-exceeded reduce would not exist. lim x ? 0? of ?(x) ln x Plugging in x = 0, you get 0 • –? it extremely is an indeterminate style. Rewrite the equation as a fragment and use l'Hôpital's rule: = lim x ? 0? of [ ln x / x^(–a million/2) ] Take the by-product of the numerator divided by technique of the by-product of the denominator: = lim x ? 0? of [ (a million/x) / (–a million/2)x^(–3/2) ] Simplify (be careful!): = lim x ? 0? of [ –2x^(a million/2) ] Plug in x = 0: = 0 lim x ? 0? of ?(x) ln x = 0

2016-12-03 05:45:42 · answer #2 · answered by lathem 4 · 0 0

You are doing the problem right it seems, you just need to simplify: -x/(2x^.5*ln^2(x))=-1/2*x^.5/ln^2(x)=0/inf as x->0+ which equals 0.

2007-04-03 11:25:12 · answer #3 · answered by bruinfan 7 · 0 0

I think you'll find life a little easier if you move sqrt(x) into the denominator instead.

lnx / x^(-1/2)

Using L'Hopital's you get
(1/x) / (-1/2 x^(3/2))
which simplifies to
-2 x^(1/2)

Then you can just evaluate.

2007-04-03 11:23:39 · answer #4 · answered by J2S 2 · 0 0

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