Limit approaching infinity with an unknown in the exponent  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

Limit approaching infinity with an unknown in the exponent

Postby jimmy_boots on Thu Dec 18, 2008 3:27 pm

Find the limit, as x approaches infinity, of (1+(2/x))x

This is a problem from Temple University's 'Calculus on the Web' site, I am trying to get re-acquainted with first-semester Calculus as I will begin taking Calc II in a month from now. I can't think of how to bring that x down. Logs?

Thanks,

Jimmy
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Re: Limit approaching infinity with an unknown in the exponent

Postby DAiv on Sat Dec 20, 2008 10:45 am

jimmy_boots wrote:Find the limit, as x approaches infinity, of (1+(2/x))x

This is a problem from Temple University's 'Calculus on the Web' site, I am trying to get re-acquainted with first-semester Calculus as I will begin taking Calc II in a month from now. I can't think of how to bring that x down. Logs?


Do you need to bring it down?

If you plug infinity into x, starting from inside the parentheses and working outwards, what do you end up with?

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Re: Limit approaching infinity with an unknown in the exponent

Postby jimmy_boots on Sat Dec 20, 2008 2:55 pm

2 divided by infinity goes to 0. 1 plus 0 is one. One raised to infinity is an indeterminate form, hence my wish to change the nature of the exponent x.
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Re: Limit approaching infinity with an unknown in the exponent

Postby DAiv on Sat Dec 20, 2008 3:44 pm

jimmy_boots wrote:2 divided by infinity goes to 0. 1 plus 0 is one. One raised to infinity is an indeterminate form, hence my wish to change the nature of the exponent x.



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Re: Limit approaching infinity with an unknown in the exponent

Postby jimmy_boots on Mon Dec 22, 2008 7:57 pm

That's not helping me dave. It looks good intuitively but 1 is not a valid answer. I repeat, one raised to infinity is considered an indeterminate form. Maybe someone else would like to chime in?
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Re: Limit approaching infinity with an unknown in the exponent  TOPIC_SOLVED

Postby DAiv on Tue Dec 23, 2008 3:32 pm

jimmy_boots wrote:That's not helping me dave. It looks good intuitively but 1 is not a valid answer. I repeat, one raised to infinity is considered an indeterminate form.

You're right, it's not a simple solution.

Q. Find

There are at least two ways of solving this, a long, drawn out way, and a much simpler way.


The long, drawn out method

You need to use L'Hôpital's rule (twice) to convert the indeterminate forms into determinate ones. This involves converting the limit to a fractional form and deriving.

Basically, you need to:

  1. Set the whole limit equal to a variable, say, y
  2. Take the natural logarithm of both sides
  3. Regroup, so the limit is out front
  4. Rearrange the natural logarithm to bring down the power
  5. Divide through to get a term containing 'x' on both top and bottom
    (This gives a limit with the form = indeterminate)
  6. Use L'Hôpital's rule to circumvent this indeterminate form
    (The limit will still have a term containing 'x' on both top and bottom, giving a limit with the form = indeterminate)
  7. Use L'Hôpital's rule again to circumvent this last indeterminate form
  8. Finally, solve for y


Here's the solution:

Spoiler:
Q. Find


[1.] Set the whole limit equal to y,



[2.] Take the natural logarithm of both sides,



[3.] Since is a continuous function, we can swap the positions of the limit and the natural logarithm,



[4.] Rearrange the natural logarithm to bring down the power,



[5.] Divide through to get a term containing 'x' on both top and bottom,



This ultimately resolves to , so...

[6.] Use L'Hôpital's rule to circumvent this indeterminate form,

a) Take the derivative of the numerator,



b) Take the derivative of the denominator,



c) Divide the derivative of the numerator by the derivative of the denominator,



Substitute the result back into our limit,



This ultimately resolves to , so...

[7.] Use L'Hôpital's rule again to circumvent this last indeterminate form,



[8.] Solve for y,




Therefore, .



The simpler method

This uses the fact that looks very similar to our given limit, .


Here's the solution:

Spoiler:
Q. Find





Substitute for ,



But , so




Therefore, .


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Re: Limit approaching infinity with an unknown in the exponent

Postby jimmy_boots on Fri Jan 09, 2009 9:25 pm

That's what I'm talking about! Thanks.

I follow the L'Hopital's Rule method of solving this problem, but where did you get the "" from?

This is a given identity thinger I should be familiar with? I haven't come across it, please let me know the deal.

Thanks again,

Jimmy
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Re: Limit approaching infinity with an unknown in the exponent

Postby DAiv on Sat Jan 10, 2009 8:53 am

jimmy_boots wrote:I follow the L'Hopital's Rule method of solving this problem, but where did you get the "" from?

This is a given identity thinger I should be familiar with? I haven't come across it, please let me know the deal.


This is just one of many ways of representing . It would be unreasonable to learn them all by heart, but a familiarisation with their general forms may alert you to the possibility of hiding in a particular problem. Learning a few of the simpler ones, like this, may prove useful (and save a lot of time).

In fact, by using the more general version of this representation of , the second solution can be simplified further:


Q. Find


Since ,


Therefore, .



Thinking about where this question originally came from, Temple University's 'Calculus on the Web' site, and the simplicity of the other 22 questions leading up to this one, I would imagine that the question may be testing for knowledge of this very representation of .



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