Find the limit as x approaches infinity of $\frac{x^{ln 2}}{1 + ln x}$ $(x^(ln2))/(1+lnx)$ ?

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Find the limit as x approaches infinity of $\frac{x^{ln 2}}{1 + ln x}$ $(x^(ln2))/(1+lnx)$ ?

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Answer 1 · 2014-09-03T10:52:32

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.
$lim_{x to infty}{x^{ln2}}/{1+lnx} =lim_{x to infty}{(ln2)x^{ln2-1}}/{1/x} = (ln2)lim_{x to infty}{x^{ln2}x^(-1)}/{x^-1}$

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

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Find the limit as x approaches infinity of $\frac{x^{ln 2}}{1 + ln x}$ $(x^(ln2))/(1+lnx)$ ?

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