What is the limit of (x - ln x) as x approaches oo?

1 Answer
Jan 22, 2017

This limit goes unbounded to +infty.

Explanation:

Trying to "substitute", gives us the indeterminate form infty - infty.

Recall that x = lne^x and lna - lnb = ln(a/b):

lim_(x->infty) (x - lnx) = lim_(x->infty) (lne^x - lnx)

=lim_(x->infty) ln(e^x/x). Let u = e^x/x.

Now we will take the limit of u as x approaches infinity:

lim_(x->infty) e^x/x. Since we get the indeterminate form infty/infty, and e^x and x

are differentiable everywhere, we can use L'Hospital's rule:

lim_(x->infty) (e^x/x) = lim_(x->infty) ((e^x)') / ((x)')

=lim_(x->infty) (e^x) = infty.

Now we know that as x -> infty, u -> infty, and

lim_(u->infty) (lnu) = infty.

Therefore,

lim_(u->infty) (lnu) = lim_(x->infty) ln(e^x/x)

=lim_(x->infty) (lne^x - lnx) = lim_(x->infty) (x - lnx).