Math Is Fun Forum

chriss

Member

Registered: 2009-02-08

Posts: 4

limits

lim (x[sup]χ[/sup][sup]1/2[/sup] ) /  3[sup]χ[/sup],
x->+infinity
can you help me to solve this?

coffeeking

Member

Registered: 2007-11-18

Posts: 44

Re: limits

Bump this up since I am interested in knowing too...

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: limits

Have you tried L'Hospital's rule?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

janetpc2008

Member

Registered: 2009-02-03

Posts: 7

Re: limits

Does it help if you just look at squares of 3? This way the fraction can be combined.
Define a subset of x:

So now we have 3 to a power in the numerator and the denominator, so we can subtract them to get the fraction:

In the limit of y going to +infinity, is this easier to manage?

I'm not sure how to prove this goes to zero, but it does.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: limits

If you can prove that

as

, then

.

Have you tried L'Hôpital's rule?

I did, and got stuck.

Last edited by JaneFairfax (2009-02-23 03:58:01)

Eigma

Member

Registered: 2009-02-22

Posts: 5

Re: limits

Hi. I'm really confused how to start proving sequences. Can you give me a technique, for example, how to prove that if an approaches A and bn approaches B, then an/bn approaches A/B where bn and B is not 0.

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"A smile is a curve that can set things straight."

LampShade

Member

Registered: 2009-02-22

Posts: 23

Re: limits

Thanks for the tip about /displaystyle

Last edited by LampShade (2009-02-23 06:06:38)

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--  Boozer

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: limits

This happens when you use $$. Use \displaystyle to restore settings.

\mbox{Let } y = \frac {x^{\sqrt{x}}}{3^x} \\
\ln{y} =\sqrt{x} \ln{x} - x \ln{3}  \\
\ln{y} = x ( $$\frac {\ln{x}}{\sqrt{x}}$$ - \ln{3} ) \\
\\
\mbox{Use L'Hospital's rule to show that} \displaystyle\lim_{x \to \infty} \mbox {of the first term is zero, and so the overall limit goes to } -\infty \mbox{.  Hence the limit you seek is 0.}