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

3 Answers
Mar 24, 2015

lim_(xrarroo)lnx=oo

To see this, we'll use:

lnx=int_1^x1/tdt

and

int_a^bf(t) dt = int_a^c f(t) dt + int_c^bf(t) dt

and

If, on [a, b] we have f(t)>=m, then int_a^b f(t)dt >=(b-a)*m

We will look at intervals of the form: [2^n, 2^(n+1)]

On [1, 2], we have 1/t >= 1/2, so
int_1^2 1/tdt >= (2-1)*1/2=1/2
And so, ln2 >= 1/2

On [2, 4], we have 1/t >= 1/4, so
int_1^4 1/tdt=int_1^2 1/tdt+int_2^4 1/tdt >= 1/2+(4-2)*1/4=1/2+1/2=1
And, ln 4 >= 2/2 =1

.

On each [2^n, 2^(n+1)], we have 1/t >= 1/(2^(n+1) So the additional integral int_(2^n)^(2^(n+1)) 1/t dt adds more than

(2^(n+1)-2^n) * 1/(2^(n+1)) = [2^n(2-1)] * 1/2^(n+1)=(2^n)/2^(n+1)=1/2

And so, ln (2^(n+1)) = int_1^(2^(n+1)) 1/t dt >= (n+1)/2

So, as xrarroo, we have int_1^x 1/t dt rarr oo.

Since this integral is ln x, we have lim_(xrarroo)lnx=oo

May 3, 2015

The answer is +infty

You can prove it by reductio ad absurdum.

You know that if x>1 ln(x)>0 so the limit must be positive.
You also know that ln(x_2)-ln(x_1)=ln(x_2/x_1) so if x_2>x_1 the difference is positive, so ln(x) is always growing

If lim_{x->infty}ln(x) = M in RR you have ln(x)< M => x < e^M, but x->infty so M can not be in RR, and the limit must be +infty

Jul 28, 2017

For any strictly increasing function, f, if a < b, then f(a) < f(b).
Let M be an arbitrary positive number.
Since (as others have shown), f(x) = lnx is strictly increasing on (0, oo),
lnx > M
if and only if
x > e^M.

Since the numbers themselves increase without bound, we have shown that by making x large enough, we may make f(x) = lnx as large as desired.

Thus, the limit is infinite as x goes to oo.