Question

Use L'Hospital Rule

Limx->infin (x-lnx)

Answer 1

\[\lim_{x \rightarrow \infty} (x-lnx)\]

Answer 2

Hmm so we're getting an indeterminate form of:\[\Large \infty - \infty\]Getting it into a form where we can apply L'Hopital is a little tricky.
Let's try exponentiating.

Answer 3

\[\LARGE \lim_{x \rightarrow \infty} (x-\ln x) \quad=\quad e^{\ln\left[\lim_{x \rightarrow \infty} (x-\ln x)\right]}\]

Answer 4

Pass the log into the limit, and let's ignore the e for right now,
This is the part we want to deal with:\[\Large \lim_{x\to\infty}\ln\left[x-\ln x\right]\]

Answer 5

Applying rules of logs allows us to write it like this:\[\Large \lim_{x\to\infty}\frac{\ln [x]}{\ln\left[\ln x\right]}\]

Answer 6

And from here, it's in a form where we're allowed to apply L'Hopital, yay!

Was that earlier step confusing? Where I wrote it with a base of e and the natural log thingy?

Answer 7

Yes, I am totally confused. I will write it out and see if It helps me understand it better.