Question

Why is
`lim_(x \to \infty) (x)/(ln x)` equal to infinity, but
`\lim_(x \to -\infty)(2^x)/(x^2)` is equal to 0?

Answer 1

The first limit is infinity since
`x>\ln(x)` for all
`x>0`.

The second limit is zero since
`2^x` converges to 0 as
`x\to-\infty`, while
`x^2` is very large and positive.