Question

How do you find the Limit of `ln(x) / sqrtx` as x approaches infinity?

Answer 1

`ln(x)` grows more slowly than any positive power of `x` as `x->oo`.

Hence the limit is `0`

Explanation:

Quick answer

Note that the exponential function `e^x` grows faster than any polynomial, in particular `x^2`

As a result the inverse function `ln x` grows more slowly than the inverse function `sqrt(x)`

Here's `e^x` and `x^2` ...

graph{(y-e^x)(y-x^2) = 0 [-39.96, 40.04, -164.4, 235.6]}

Here's `ln x` and `x^(1/2)` ...

graph{(y - ln x)(y - x^(1/2)) = 0 [-96.4, 100.36, -4.56, 5.44]}

Hence:

`lim_(x->oo) (ln x)/(sqrt(x)) = 0`