Limit exponential/power function

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Just Someone
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Joined: Sun Nov 24, 2013 8:05 pm UTC

Limit exponential/power function

Postby Just Someone » Sun Nov 24, 2013 8:21 pm UTC

Hey :)

I came across the following question in my homework: determine the limit for x to 3 of (x/3)1/(x-3). The previous questions about limits where either solvable by l'Hôpital or Taylor series, but this one seems quite different. If I rewrite it as a fraction, the dividing by 0 still remains. And for the Taylor series, I have no idea how that should be done. I can write something like ax in Taylor form, but this seems quite impossible and I don't know how I should continue if I manage to rewrite it. Maybe it is a combination of the two, but I think that I'm going off track if I do that. Another thing I tried was writing it as an exponential function (eln(x/3)...), but that wasn't useful either

So I'm stuck, I hope you guys could give me a hint about how to solve this thing, because I don't get it.. :P

skullturf
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Re: Limit exponential/power function

Postby skullturf » Sat Nov 30, 2013 7:03 pm UTC

Just Someone wrote:Another thing I tried was writing it as an exponential function (eln(x/3)...), but that wasn't useful either


That should work. Personally, the way that I prefer to write these types of limits is as follows:

Let y = (x/3)1/(x-3) (which of course is of the form a^b where a and b both depend on x)

Then ln y = (1/(x-3)) ln(x/3)

It should be possible to find the limit of ln y (as x approaches 3) using fairly standard methods.

flownt
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Re: Limit exponential/power function

Postby flownt » Mon Dec 02, 2013 3:23 pm UTC

and of course ln/exp is continuous so the answer translates.

skullturf
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Re: Limit exponential/power function

Postby skullturf » Mon Dec 02, 2013 8:30 pm UTC

Yes. So, for example, if ln(y) approached 15 (it doesn't, I just made that up) then y would approach e^15.


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