6 tutors & still can't solve a limit problem (calculus)

[BehindEnemyLines]

I've spent more than four hours, asked six tutors, and still can't find an answer.

The instruction: Use L'Hospital's Rule to find the limit of:

500( 1+ 0.05/x )^(10x) as x --> infinity

I (& the tutors tried) to use the natural log, like so: let y = 500( 1 + 0.05/x )^(10x), then
ln y = ln 500( 1 + 0.05/x )^(10x)
=> ln y = (10x) ln 500( 1 + 0.05/x )
=> lim[ ln y ] = lim[ (10x) ln 500( 1 + 0.05/x ) ] as x --> infinity

Answer below (I just don't know how the calculator got that):

Answer: 500 * sqrt(e)

 

[fsstrike]

Just take the given route and multiply it with the calculation of the membrane but dont forget to find the particualr mass and negative overload factor.

 

[MrScott81]

Originally posted by: fsstrike
Just take the given route and multiply it with the calculation of the membrane but dont forget to find the particualr mass and negative overload factor.

dick

 

[LordJezo]

It's called L'Hospital's rule because after trying to use it you will end up in el'Hospital!!

 

[Dissipate]

Originally posted by: scottdog81

Originally posted by: fsstrike
Just take the given route and multiply it with the calculation of the membrane but dont forget to find the particualr mass and negative overload factor.

dick

He's not a dick, he's a yuk yuk.

 

[AgentEL]

Originally posted by: BehindEnemyLines

I (& the tutors tried) to use the natural log, like so: let y = 500( 1 + 0.05/x )^(10x), then
ln y = ln( 1 + 0.05/x )^(10x)
=> ln y = (10x) ln( 1 + 0.05/x )
=> lim[ ln y ] = lim[ (10x) ln( 1 + 0.05/x ) ] as x --> infinity

Answer below (I just don't know how the calculator got that):

Answer: 500 * sqrt(e)

What happened to the "500" term in your "ln y" function?

 

[BehindEnemyLines]

I'm more interested in HOW the calculator got the answer though.

CORRECTED. I left out the "500".

 

[bonkers325]

the calculator has deceived you. work it out by hand and check the answer in the back of the book or with the instructor.

dont plug it into ur ti-89. graph it and see how the function looks like over time. it's either going to be 500 or infinity.

 

[MrScott81]

actually just looking at the problem i would have guessed that the answer was 500, not 500*sqrt(e)...just plug in infinite to the equation:

500(1+0.05/x)^(10x) as x goes to inf
= 500(1+0.05/inf) ^ (10*inf)
= 500(1+0)^(inf)
= 500(1)
= 500

 

[BehindEnemyLines]

The calculator is the TI-89 with limit capability. I also used Mathematica (from Wolfram Research Inc).
The calculator gives me 500 * sqrt(e) and Mathematica gives me an approximation of it as ~824.361

The L'Hospital Rule requires that you change the original function so that it's in the form of:

lim[ f(x) / g(x) ] = lim[ f '(x) / g '(x) ] as x --> c. (the right limit is f prime and g prime)

But the limits of the functions f(x) and g(x) MUST be infinity / infinity OR zero / zero.

These are indeterminate form of limits (you can't tell whether a limit exists):

0*infinity, 1^infinity, infinity - infinity, 0^0, infinity^0

 

[MrScott81]

you're not using L'Hopitals rule though, so why are you trying to do log stuff?

 

[BehindEnemyLines]

I'm trying to put so that I've the limit of the numerator and denominator to be infinity / infinity or zero / zero. After that, I can apply L'Hospital's Rule.

If we take the limit as it of 500( 1+ 0.05/x )^(10x) as x --> infinity, then it has the form of 1^infinity. Which is indiscrimant, unless I missed something here.

 

[AgentEL]

Originally posted by: BehindEnemyLines
I'm trying to put so that I've the limit of the numerator and denominator to be infinity / infinity or zero / zero. After that, I can apply L'Hospital's Rule.

If we take the limit as it of 500( 1+ 0.05/x )^(10x) as x --> infinity, then it has the form of 1^infinity. Which is indiscrimant, unless I missed something here.

Would it be easier if you rearranged it to:

y = 500((x+0.05)/x))^10x

 

[RaynorWolfcastle]

ok, I'll post the solution, give me a minute to type it out.

OK here it is, the longwinded way:
y = limit(500*(1+.05/x)^10x)
y/500 = limit(1+.05/x)^10x
ln(y/500) = limit(10x*ln(1+.05/x)) => 0*infinity so we rewrite
ln(y/500) = limit(ln[(1+.05/x)/(1/10x)] => 0/0 so we use l'hopital I'll skip the limit() on the steps below for convenience
= 1/(1+.05/x)*(-.05/x^2) / (-1/10x^2)
we rewrite the numerator as 1/(x(20x+1) and cancel the negative signs to get
= [1/(x(20x+1)] / [1/10x^2] which we can now rewrite in a more readable form
= 10x^2/[x(20x+1)]
= 10x/(20x+1), this limit as x -> infinity is easy
= 1/2

so now we have
ln(y/500) = 1/2
which we rewrite as
y = 500*sqrt(e) ~=824

Your tutors suck for not figuring it out btw.

 

[Savij]

Does this help:

ln y = (10x) ln 500( 1 + 0.05/x )
= (ln 500(1+0.05/x))/(10x)^-1

 

[bleeb]

That problem is HELLA easy dawg!

 

[BehindEnemyLines]

I've tried re-arranging the function in so many different ways, but just couldn't get it into the proper form.

 

[jurzdevil]

the 500 is just a coeffcient so you can take it outside the limit.

 

[bonkers325]

carry the 500 out, you get 500 lim 10x * [ln(1+.05/x)] x -> inf

this will become inf * 0 which u will then apply l'hopitals rule to.

 

[JohnCU]

Originally posted by: scottdog81
actually just looking at the problem i would have guessed that the answer was 500, not 500*sqrt(e)...just plug in infinite to the equation:

500(1+0.05/x)^(10x) as x goes to inf
= 500(1+0.05/inf) ^ (10*inf)
= 500(1+0)^(inf)
= 500(1)
= 500

You can't just plug it in, because that's a 1^infinity type of indeterminate.

 

[PowerMacG5]

I am not going to help you simply for the egregious error of continuing to say "L'Hospital" after numerous people said "L'Hopital."

 

[JohnCU]

Originally posted by: Marauder911
I am not going to help you simply for the egregious error of continuing to say "L'Hospital" after numerous people said "L'Hopital."

Um, that's the correct way to spell it.

 

[PowerMacG5]

Originally posted by: JohnCU

Originally posted by: Marauder911
I am not going to help you simply for the egregious error of continuing to say "L'Hospital" after numerous people said "L'Hopital."

Um, that's the correct way to spell it.

Uhm, It's L'Hopital, not L'Hospital. The OP continued to say L'Hospital.

 

[JohnCU]

Originally posted by: Marauder911

Originally posted by: JohnCU

Originally posted by: Marauder911
I am not going to help you simply for the egregious error of continuing to say "L'Hospital" after numerous people said "L'Hopital."

Um, that's the correct way to spell it.

Uhm, It's L'Hopital, not L'Hospital.

I guess my book and my teacher spell it wrong then.

 

[bonkers325]

it's l'hopital