Lim(x-lnx) as x approaches infinity?

Question

I have to use L'Hopital's Rule to solve this. The answer is obviously infinity, but I do not know how to rearrange the problem to get either 0/0 or infinity/infinity.

Answer 1

Start by

e^ (lim x-> infinity)  = e ^ (x - ln(x))

By exponential properties you then have

e^(lim x-> infinity) = e^x / x.

Then use l'hospitals Rule.

Answer 2

lim (x-lnx)/x = 1 - lim lnx/x = 1 + 1/x^2 = 1 as x->infinity

So, lim (x-lnx) has the same order to approach infinity as lim x as x->infinity.

Answer 3

lim [ x - ln(x) ]

x -> infinity

As you already know, this is in the form infinity - infinity, so we have to change its form.

One thing you can do is change the form of x to ln(e^x), because ln and e are inverses of each other and x = ln(e^x).

lim [ ln(e^x) - ln(x) ]

x -> infinity

Now, we can combine the logs as per the identity.

lim [ ln( e^x / x ) ]

x -> infinity

And we can move the limit inside the log.

ln [ lim (e^x / x ) ]

. .  . x -> infinity

And now we can apply L'Hospital's rule.

ln [ lim ( e^x  /  1 ) ]

. . . x -> infinity

ln [ lim ( e^x  ) ]

. . . x -> infinity

As x approaches infinity, e^x approaches infinity.

As e^x approaches infinity, so does ln(e^x).

Therefore, the answer is infinity.