Limit of an Exponential Function

[BraedenP]

Homework Statement

\lim_{x \to \infty} \left(e^x-x \right )^{1/x}

Homework Equations

The Attempt at a Solution

Should I be equating this to f(x), taking the log of both sides, using L'Hopital's rule on the resulting indeterminate quotient? If I do that, then I end up with a perpetual indeterminate quotient no matter how many times I use L'Hopital's rule because of the e^x in the numerator.

How should I go about doing this?

 

[lineintegral1]

Yes, use L'Hospitals. Your idea is absolutely correct. I'm a little confused though as to where you got the perpetual quotient. Can you show us where this occurs? The limit should end up being e (because the limit of the logged side is 1).

 

[BraedenP]

You know what? I'm not quite sure how I got it...

But I just tried it again and it worked perfectly fine.

Thanks!

 

[HallsofIvy]

Why are you only takinng the derivative of the numerator?

L'Hopital's rule says that if f(a)= g(a)= 0, then
\lim_{x\to a}\frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}

Here,
f'(x)= \frac{e^x- 1}{e^x- x}
and
g'(x)= 1.

You should be finding
\lim_{x\to\infty}\frac{e^x- 1}{e^x- x}
and that is easy- divide both numerator and denominator by e^x.

 

[dextercioby]

Assuming the limit exists and is equal to A, then

\ln A = \ln \lim_{x \to \infty} \left(e^x-x \right )^{1/x} = \lim_{x \to \infty} \left\frac{1}{x} \ln \left[e^{x}\left(1-\frac{x}{e^x}\right)\right] = ...

You can now use the properties of the \ln function.

 

[BraedenP]

Why are you only takinng the derivative of the numerator?

L'Hopital's rule says that if f(a)= g(a)= 0, then
\lim_{x\to a}\frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}

Here,
f'(x)= \frac{e^x- 1}{e^x- x}
and
g'(x)= 1.

You should be finding
\lim_{x\to\infty}\frac{e^x- 1}{e^x- x}
and that is easy- divide both numerator and denominator by e^x.

Yeah, I realized shortly after I posted that I forgot to take the derivative of the denominator.

Thanks. :)