Question

What is the limit as x approaches infinity of `x^(ln2)/(1+ln x)`?

Answer 1

`lim_{x \to +oo} \frac{x^{ln 2}}{1+ln x} = +oo`

This can be found by using L'Hospital's Rule, which states that every limit of a fraction is equal to the limit of the derivatives of the fraction. More formally:

`lim \frac{f(x)}{g(x)} = lim \frac{f'(x)}{g'(x)}`

Using this rule on this problem:
`lim_{x \to +oo} \frac{x^{ln 2}}{1+ln x} =`
# lim_{x \to +oo} \frac{ln (2)*x^{(ln 2)-1}}{\frac{1}{x}} = lim_{x \to +oo} ln(2)*x^{ln(2)-1}*x#
`lim_{x \to +oo} ln(2)*x^{(ln 2)-1+1}=lim_{x \to +oo} ln(2)*x^{(ln 2)}`
Now all you have to know is that when `x` goes to infinity, `x^{ln2}` also goes to infinity.