Question

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

Answer 1

`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`

Answer 2

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`

Answer 3

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`.