Limit, x->0, of [ln(e^x - x)]/x

[figarcia]

Can someone help me obtaining the following limit without using l'Hopital rule?

lim_x->0(ln(e^x-x)/x)

Thanks!!!

 

[DrPhil]

Can someone help me obtaining the following limit without using l'Hopital rule?

lim_x->0(ln(e^x-x)/x)

Thanks!!!

What have you tried - it would help us to see your work!

e^x --> 1, so the numerator approaches ln(1-x). Do you know the Taylor expansion for that?

 

[lookagain]

e^x --> 1, so the numerator approaches ln(1-x).

DrPhil,

don't you mean that the numerator approaches \(\displaystyle \ln(1 - 0), \ \) because x is approaching 0?

 

[Bob Brown MSEE]

e^x --> 1, so the numerator approaches ln(1-x). Do you know the Taylor expansion for that?

Unfortunately ln(e^x-x) is not approximated by ln(1-x) near the origin. They have different first d/dx.
L'Hôpital's rule works well for this problem.

 

[Bob Brown MSEE]

Unfortunately ln(e^x-x) is not approximated by ln(1-x) near the origin. They have different first d/dx.

However, ln(e^x-x) is approximated by ln((1+x+xx/2)-x) near the origin.

Your idea might work with Taylor expansion of ln(1+xx/2) divided by x.

 

[figarcia]

However, ln(e^x-x) is approximated by ln((1+x+xx/2)-x) near the origin.

Your idea might work with Taylor expansion of ln(1+xx/2) divided by x.

Hi! Thanks for the replies. I think it works with the Taylor expansion. If you have other suggestions, I would be grateful to hear them.

Filo

 

[figarcia]

DrPhil,

don't you mean that the numerator approaches \(\displaystyle \ln(1 - 0), \ \) because x is approaching 0?

I think you are right. The l'Hopital rule works well, however, I am asked to solve it without. I guess that using the Taylor expansion is the appropriate solution. Any other suggestions?

Filo

 

[DrPhil]

I think you are right. The l'Hopital rule works well, however, I am asked to solve it without. I guess that using the Taylor expansion is the appropriate solution. Any other suggestions?

Filo

Using the Taylor expansion is a way to prove l'Hosptal's Rule, so using it is a good plan.

Bob Brown MSEE did the expansion right:

e^x - x --> (1 + x/1! + x^2/2! + ...) - x = 1 + x^2/2! + . . .

And the first term of the expansion of the ln(1+delta) = . . .

 

[Bob Brown MSEE]

need Taylor expansion of f(x)=ln(1+xx/2)
note:
f(0)=0
f'(0)=0
f''(0)=1
f(x)~xx/2

 

[Bob Brown MSEE]

ln(e^x-x)/x < x
ln(e^x-x) < xx
ln(e^x-x) - xx <0
can perhaps be shown

ln(e^x-x)/x > 0

and squeeze to 0.

 

[lookagain]

ln(e^x-x)/x < x \(\displaystyle \ \ \) for x > 0 *
ln(e^x-x) < xx \(\displaystyle \ \ \) except at x = 0 **
ln(e^x-x) - xx <0 \(\displaystyle \ \ \) See ** above.
can perhaps be shown

ln(e^x-x)/x > 0 \(\displaystyle \ \ \) See * above.

and squeeze to 0.

.

 

[DrPhil]

need Taylor expansion of f(x)=ln(1+xx/2)
note:
f(0)=0
f'(0)=0
f''(0)=1
f(x)~xx/2

After this step we can just take the limit:

\(\displaystyle \displaystyle \lim_{x \to 0} \dfrac{\ln(e^x - x)}{x} = \lim_{x \to 0} \dfrac{x^2/2}{x} = \lim_{x \to 0} \dfrac{x}{2} = 0\)

Nothing requires \(\displaystyle x>0\), so the limit exists approaching 0 from either direction.