L'Hopital's Rule to find limit

[Peapod]

I understand the basics of l'Hopital's Rule, but I'm having trouble figuring out how to get started with this question, especially since it's a difference rather than a quotient:

lim x->inf [x - ln(x^3 - 1)]

The question also gives the hint: "ln e^x = x", but I'm unsure how to apply it.

If anyone could help me get pointed in the right direction with this problem, I'd be very grateful.

 

[ksdhart]

The trick here is that you don't know the value of the limit, but you know it does have some value L. So, we have the following declaration:

\(\displaystyle \displaystyle \lim _{x\to \infty }\left(x-ln\left(x^3-1\right)\right)=L\)

Then, what if we try to undo the natural log?

\(\displaystyle \displaystyle e^{\lim _{x\to \infty }\left(x-ln\left(x^3-1\right)\right)}=e^L\)

And there's a property involving e to the power of a limit:

\(\displaystyle \displaystyle e^{\lim _{x\to C}\left(f\left(x\right)\right)}=\lim _{x\to C}\left(e^{f\left(x\right)}\right)\)

So, applying that property we get:

\(\displaystyle \displaystyle e^L=\lim _{x\to \infty }\left(e^{x-ln\left(x^3-1\right)}\right)\)

Try proceeding from there and see what you get.

 

[Ishuda]

I understand the basics of l'Hopital's Rule, but I'm having trouble figuring out how to get started with this question, especially since it's a difference rather than a quotient:

lim x->inf [x - ln(x^3 - 1)]

The question also gives the hint: "ln e^x = x", but I'm unsure how to apply it.

If anyone could help me get pointed in the right direction with this problem, I'd be very grateful.

The way I would do it:
x - ln(x³-1) = ln(e^x) - ln(x³-1) = ln(\(\displaystyle \frac{e^x}{x^3-1}\))
Thus
\(\displaystyle \underset{x \to \infty}{lim}[ x - ln(x^3-1)]\,=\, \underset{x \to \infty}{lim}[ ln(\frac{e^x}{x^3-1}) ]\, =\, ln(\underset{x \to \infty}{lim}[\frac{e^x}{x^3-1}])\)
...