Question

Find the limit as x approaches infinity of `(x^(ln2))/(1+lnx)`?

Answer 1

By l'Hopital's Rule,
`lim_{x to infty}{x^{ln 2}}/{1+lnx}=infty`.

Intuitively, Imagine that this is a fight between the numerator and the denominator, and `x^{ln2}` grows faster than `1+lnx` as x approaches `infty`, which means that the numerator will be larger than the denominator; therefore, the quotient tends to infinity.

Let us use l'Hopital's Rule to find the limit.
#lim_{x to infty}{x^{ln2}}/{1+lnx} =lim_{x to infty}{(ln2)x^{ln2-1}}/{1/x} = (ln2)lim_{x to infty}{x^{ln2}x^(-1)}/{x^-1} #

`=(ln2) lim_{x to infty}x^{ln2}=infty`