Question

How do you find the limit of `lnx/lnx^2` as `x->oo`?

Answer 1

Use the properties of logarithms to rewrite `lnx^2 = 2lnx`

Explanation:

So, `lnx/(lnx^2) = lnx/(2lnx) = 1/2`.

So the function is a constant `f(x) = 1/2` and the limit at infinity is `1/2`