Question

What is the limit of `(lnx)^2/x` as x approaches `oo`?

Answer 1

`0`

Explanation:

`(log_ex)^2/x = ((log_e x)/sqrt(x))^2` so
`lim_{x->oo}(log_ex)^2/x = (lim_{x->oo}(log_e x)/sqrt(x))^2` but
`lim_{x->oo}(log_e x)/sqrt(x)=lim_{x->oo}(e^{log_e x})/e^{sqrt(x)} = lim_{x->oo}(x)/e^{sqrt(x)}`

Making now `sqrt(x) = y`

`lim_{x->oo}(x)/e^{sqrt(x)}equivlim_{y->oo}(y^2)/e^y = 0`