Question

How do you find `\lim _ { x \rightarrow + \infty } \frac { x } { \ln ( 1+ 3x ^ { 2} ) }`?

Answer 1

I got that it tends to `oo`.

Explanation:

If we solve directly we should get the indeterminate form `oo/oo`.
However, we can use de L'Hospital Rule deriving separately top and bottom and THEN apply the limit:
`lim_(x->oo)1/(1/(1+3x^2)*6x)=lim_(x->oo)(1+3x^2)/(6x)`
Collect `x`:
`lim_(x->oo)(cancel(x)(1/x+3x))/(6cancel(x))=oo`
As when `x->oo` we have that `1/x->0`