comparison of two functions

PUBLISHED: MAY 2, 2026•1 MIN READ

Compare the growth of two functions0 implies has smaller order of growth than c implies has same order of growth as implies has larger order of growth than

Abhishek SinghAuthor

azurill

Compare the growth of two functions

\lim_{n\to\infty}\frac{t(n)}{g\left(n\right)}

0 implies has smaller order of growth than
c implies has same order of growth as
implies has larger order of growth than

Compare the growth of the function and

let
t(n) =
g(n) =

\lim_{n\to\infty}\frac{t(n)}{g(n)}=\lim_{n\to\infty}\frac{nlog_2n}{4^{n}}

\frac{\infty\log_2 \infty}{4^\infty}=\frac{\infty}{\infty}

This is an form, so apply L'Hopital’s Rule (differentiate numerator and denominator):

Apply L hopital Rule

According to L Hopital can only be applied in the case where direct substitution yields an indeterminate form, meaning or .

\lim_{n\to\infty}\frac{\frac{d}{dn}(nlog_2n)}{\frac{d}{dn}4^{n}}

\frac{d}{dn}\left(nlog_2n\right)

\frac{n\frac{d}{dn}(log_2n)+\frac{d}{dn}(n)\cdot log_2n}{\frac{d}{dn}\left(4^{n}\right)}

Convert to natural log:

\frac{d}{dn}\frac{\ln{n}}{\ln2}

\frac{1}{\ln{2}}\cdot\frac{d}{dn}\log n

\frac{1}{\ln{2}} \cdot \frac{1}{n} = \frac{1}{n\ln{2}}

\boxed{\frac{d}{dn} \log_2{n} = \frac{1}{n\ln{2}}}

\boxed{\frac{d}{dn} 4^n = 4^nln4}

Put the value

\frac{n\cdot\frac{1}{n\ln{2}}+1\cdot log_2n}{4^{n}\ln4}

\lim_{n\to\infty}\frac{\left(\frac{1}{\ln{2}}+\log_2n\right)}{4^{n}\ln{4}}

Again, apply L hopital Rule

Differentiate numerator

\frac{d}{dn}\left(\frac{1}{\ln2}+\log_2n\right)

0+\frac{1}{n\ln{2}}=\frac{1}{n\ln2}

Differentiate denominator

\ln4\cdot\frac{d}{dn}4^{n}

\ln{4}\cdot4^{n}\ln4

(\ln{4})^2 \cdot 4^n

Put the values:

\frac{1\text{/}n\ln2}{(\ln{4})^2\cdot4^{n}}

\lim_{n\to\infty}\quad\frac{1}{(n\ln2)\cdot(\ln{4})^2\cdot4^{n}}

= \frac{0}{\infty} = 0

\boxed{\text{0 implies t(n) has smaller order of growth than g(n)}}

t(n)=n\log_2n

g(n) = 4^n