The Student Room Group

Having trouble with a limit

Hi,

I'm stuck on finding this limt: limx0xnex1ex \lim_{x\rightarrow 0}\frac{x^{n}e^{x}}{1-e^{x}} .

I've tried what I think to be every approach (L'hopital's, L'hopitals with induction, sandwich rule, manipulating the function etc.) but have had no success.

Does anyone know how to go about solving this?

Thanks,
Dan
Reply 1
Original post by DanKeitley
Hi,

I'm stuck on finding this limt: limx0xnex1ex \lim_{x\rightarrow 0}\frac{x^{n}e^{x}}{1-e^{x}} .

I've tried what I think to be every approach (L'hopital's, L'hopitals with induction, sandwich rule, manipulating the function etc.) but have had no success.

Does anyone know how to go about solving this?

Thanks,
Dan


definitely L hospital for me

(and let the purists argue...)
Reply 2
Original post by TeeEm
definitely L hospital for me

(and let the purists argue...)


Could you expand a little...? haha
Reply 3
Original post by DanKeitley
Hi,

I'm stuck on finding this limt: limx0xnex1ex \lim_{x\rightarrow 0}\frac{x^{n}e^{x}}{1-e^{x}} .

I've tried what I think to be every approach (L'hopital's, L'hopitals with induction, sandwich rule, manipulating the function etc.) but have had no success.

Does anyone know how to go about solving this?

Thanks,
Dan


I think l'Hopital will work - too late in the evening for me to try this though :biggrin:

One thing to note to help yourself out a bit - note that the e^x in the numerator is irrelevant: because it tends to a finite non-zero limit as x->0 it doesn't affect whether or not the whole thing tends to a limit so you can "factor it out" and multiply it back in using the algebra of limits at the end.

You haven't said what n is, however - is it a general integer, a natural number or what? This will affect your conclusion!

EDIT: I've just thought of another way of doing it too, which is to write the e^x in the numerator in terms of the denominator and then separating things out.
Reply 4
Original post by DanKeitley
Could you expand a little...? haha



Original post by davros
I think l'Hopital will work - too late in the evening for me to try this though :biggrin:

One thing to note to help yourself out a bit - note that the e^x in the numerator is irrelevant: because it tends to a finite non-zero limit as x->0 it doesn't affect whether or not the whole thing tends to a limit so you can "factor it out" and multiply it back in using the algebra of limits at the end.

You haven't said what n is, however - is it a general integer, a natural number or what? This will affect your conclusion!

EDIT: I've just thought of another way of doing it too, which is to write the e^x in the numerator in terms of the denominator and then separating things out.


sorry but I was teaching late tonight and I typed L'hospital without much thinking, so this is what I would do

write exponentials (top and bottom) as power series

on the top the series will start with x^n plus higher order terms
on the denominator, once the 1s cancell out it will start with -x plus higher order terms

divide an x all the way through should be enough to show that the limit is in fact zero

PS I have assumed that n is positive integer/positive real
Reply 5
Original post by davros
I think l'Hopital will work - too late in the evening for me to try this though :biggrin:

One thing to note to help yourself out a bit - note that the e^x in the numerator is irrelevant: because it tends to a finite non-zero limit as x->0 it doesn't affect whether or not the whole thing tends to a limit so you can "factor it out" and multiply it back in using the algebra of limits at the end.

You haven't said what n is, however - is it a general integer, a natural number or what? This will affect your conclusion!

EDIT: I've just thought of another way of doing it too, which is to write the e^x in the numerator in terms of the denominator and then separating things out.


Thanks for your comment. n is a positive integer :smile: I couldn't get this to work but I have a tutorial session soon and will ask about this way of doing it.


Original post by TeeEm
sorry but I was teaching late tonight and I typed L'hospital without much thinking, so this is what I would do

write exponentials (top and bottom) as power series

on the top the series will start with x^n plus higher order terms
on the denominator, once the 1s cancell out it will start with -x plus higher order terms

divide an x all the way through should be enough to show that the limit is in fact zero

PS I have assumed that n is positive integer/positive real

Ahhhhh! Very clever :clap2:. Thanks very much. I wouldn't have thought of that!
Reply 6
Original post by DanKeitley
Thanks for your comment. n is a positive integer :smile: I couldn't get this to work but I have a tutorial session soon and will ask about this way of doing it.



Ahhhhh! Very clever :clap2:. Thanks very much. I wouldn't have thought of that!


I am glad it made sense. Goodnight
Original post by DanKeitley
Hi,

I'm stuck on finding this limt: limx0xnex1ex \lim_{x\rightarrow 0}\frac{x^{n}e^{x}}{1-e^{x}} .

I've tried what I think to be every approach (L'hopital's, L'hopitals with induction, sandwich rule, manipulating the function etc.) but have had no success.

Does anyone know how to go about solving this?

Thanks,
Dan


Use L'hopitals rule:

The derivative of the top number is nx^(n-1)*e^x + x^n*e^x.

The derivative of the bottom number is -e^x.

So L'hopital rules states you can instead evaluate the limit of:

lim of ([nx^(n-1)*e^x + x^n*e^x]/-e^x)

The e^x will cancel yielding:

lim of (-nx^(n-1) - x^n)

So now simply put x = 0, and we get find that the limit is 0.
(edited 9 years ago)
Reply 8
Original post by DanKeitley
Thanks for your comment. n is a positive integer :smile: I couldn't get this to work but I have a tutorial session soon and will ask about this way of doing it.



Ahhhhh! Very clever :clap2:. Thanks very much. I wouldn't have thought of that!


Note that ex=(ex)=(ex+1)+1e^x = -(-e^x) = -(-e^x + 1) + 1 so that

xnex1ex=xn(1ex)+xn1ex=xn+xn1ex\dfrac{x^ne^x}{1 - e^x} = \dfrac{-x^n(1 - e^x) + x^n}{1 - e^x} = -x^n + \dfrac{x^n}{1 - e^x}

If n is a positive integer then the limit of x^n is easy to find and you can concentrate on the 2nd fraction which is a bit easier to work with than the original expression.

Quick Reply

Latest