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0 divided by 0?

Hi,
I've recently got this question:

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be the function f(x):=ex−1−x−x22x3f(x):=\frac{e^x-1-x-\frac{x^2}{2}}{x^3}
Compute f(0)f(0).You may use l'Hôpital's rule provided that you can justify why you can apply it.

Now if i compute f(0), i.e. subsitute x in the formula for 0, as the question asks, it just simplifies to 00\frac{0}{0}. This, however is undefined. As far as l'Hôpital's rule goes, as far as i was aware, it deals with limits, rather than actual values.

This makes me believe that there's a huge typo in the question, and that it wants me to evaluate the limit as x tends to 0. Is there any other way of interpreting this question that I can't see?
Help would be appreciated, this is the last piece of coursework i've got before the holidays and I'd love to get it out of the way ASAP.

Cheers.
Reply 1
Avatar for DJ_Black
DJ_Black
black hole / Maths Error 2


the choice is yours
Reply 2
Avatar for Dekota-XS
Dekota-XS
2
The question per se doesn't make sense unless it's part of a Limits exercise. I suspect it wants you to find f(x) as x -> 0.
Reply 3
Avatar for MillerTraub
MillerTraub
Original post by immense010
Hi,
I've recently got this question:

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be the function f(x):=ex−1−x−x22x3f(x):=\frac{e^x-1-x-\frac{x^2}{2}}{x^3}
Compute f(0)f(0).You may use l'Hôpital's rule provided that you can justify why you can apply it.


Cheers.



Use the taylor expansion of the exponential term.

Boom
(edited 13 years ago)
Reply 4
Avatar for immense010
immense010
OP
Original post by MillerTraub
Use the taylor expansion of the exponential term.

Boom


Oh yeah, didn't notice that. :biggrin: Only problem is the Taylor series expansion is infinite, so although it will cancel out the other 3 terms in the numerator, I'll still be left with an infinite series, with each term containing x to a certain power, so when I try substituting in x=0 it'll still give me 0/0

I got this homework for my analysis module, so I need to give an exact answer to the question, and so I can't just use the Taylor expansion of order 2, as this would be an approximation. Although I do agree with you that it would behave as f(x)=0 sufficiently close to the origin.
Original post by immense010
Oh yeah, didn't notice that. :biggrin: Only problem is the Taylor series expansion is infinite, so although it will cancel out the other 3 terms in the numerator, I'll still be left with an infinite series, with each term containing x to a certain power, so when I try substituting in x=0 it'll still give me 0/0

I got this homework for my analysis module, so I need to give an exact answer to the question, and so I can't just use the Taylor expansion of order 2, as this would be an approximation. Although I do agree with you that it would behave as f(x)=0 sufficiently close to the origin.


o'rly?
Reply 6
Avatar for MillerTraub
MillerTraub
Original post by immense010
Oh yeah, didn't notice that. :biggrin: Only problem is the Taylor series expansion is infinite, so although it will cancel out the other 3 terms in the numerator, I'll still be left with an infinite series, with each term containing x to a certain power, so when I try substituting in x=0 it'll still give me 0/0

I got this homework for my analysis module, so I need to give an exact answer to the question, and so I can't just use the Taylor expansion of order 2, as this would be an approximation. Although I do agree with you that it would behave as f(x)=0 sufficiently close to the origin.


What is x^2 / x equal to when x=0?
Reply 7
Avatar for immense010
immense010
OP
aah lol ok i've spotted it. Thanks a lot ^^
Reply 8
Avatar for MillerTraub
MillerTraub
The moral of the story here is:

If in doubt, the answer is always 1/6

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