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

1 Answer
Dec 16, 2014

#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.