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Limit question

by redsox5
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redsox5
#1
Oct23-07, 07:03 PM
P: 40
The problem is

The limit as x approaches pos infinity ln(square root of x + 5) divided by ln(x)

In the numerator only x is under the square root. I'm having trouble getting to this answer. If someone can take a look I would really appreciate it.
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mathwonk
#2
Oct23-07, 09:35 PM
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remember ln(sqrt(x)) = 1/2 ln(x).
redsox5
#3
Oct23-07, 09:55 PM
P: 40
well you can multiply by sqr. rt of x -5/ sqr rt of x -5

that leaves you with ln x-25/ x(sqr rt of x + 5)

Gib Z
#4
Oct24-07, 05:48 AM
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Limit question

mathwonk's hint intentionally disregarded the 5 within the argument of the log. Intuitively, as x grows large the 5 within the log becomes insignificant and can be ignored. More rigorously, the natural log of (sqrtx + 5) is asymptotic to log (sqrtx), which means that the difference of the two for a given value of x goes to zero as x goes to infinity, basically [tex]\lim_{n\to\infty} \frac{ \ln (\sqrt{x} +5)}{ \ln \sqrt{x}} = 1[/tex].

If you want to take your route, you would need to multiply by the log of (sqrtx - 5) instead.
JonF
#5
Oct24-07, 06:51 AM
P: 617
if you need to formally show this do you know l'hopital's rule?
redsox5
#6
Oct24-07, 08:20 AM
P: 40
no, i don't know that rule yet. But gib, the actual problem has ln(x) in the denomator, not the sqr. rt. So does that make it 0?
redsox5
#7
Oct24-07, 06:46 PM
P: 40
well on a calculator i come up with 1/2
can someone tell me the best way to go about solving this?
Gib Z
#8
Oct25-07, 07:06 AM
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I know the denominator doesnt have the sqrt mate, but your missing my point. Im saying, The numerator can be replaced with ln(sqrtx) instead of the whole thing, because of the reasons i said before: the plus 5 becomes insignificant as x goes to infinity!! If you just neglected the 5, which you have shown you can do, take mathwonks post into account and take this problem down!!
redsox5
#9
Oct25-07, 03:45 PM
P: 40
.5 got it thanks
ZioX
#10
Oct26-07, 04:34 PM
P: 371
Did someone delete my post? ;0


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