Compare the growth of the functions

PUBLISHED: MAY 2, 2026•1 MIN READ

DAA Assignment 1Compare the growth of the functions and lett(n) = g(n) = This is an form, so apply L'Hopital's RuleApply L hopital Ruledifferentiate the numer

Divya Sachan
Divya SachanAuthor
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DAA Assignment 1

Compare the growth of the functions and

let
t(n) =
g(n) =
lim⁡n→∞t(n)g(n)=lim⁡n→∞n2nlog⁡2n\lim_{n\to\infty}\frac{t(n)}{g(n)}=\lim_{n\to\infty}\frac{n^2}{n\log_2n}

This is an form, so apply L'Hopital's Rule

Apply L hopital Rule

differentiate the numerator

t′(n)=ddn⋅n2=2nt^{\prime}(n)=\frac{d}{dn}\cdot n^2=2n

Differentiate thedinominator

t′(n)g′(n)=2nlog⁡2n+1ln⁡2\frac{t'(n)}{g'(n)}=\frac{2n}{\log_2n+\frac{1}{\ln{2}}}
lim⁡n→∞t′(n)g′(n)=∞∞\lim_{n\to\infty}\frac{t^{\prime}\left(n\right)}{g^{\prime}(n)}=\frac{\infty}{\infty}

Apply L hopital Rule, Again

t=a+bt^{}=a+b
t(n)=ddn⋅2nt^{}\left(n\right)=\frac{d}{dn}\cdot2n
t(n)=2\boxed{t^{}(n)=2}
g"(n)=ddn(log⁡2n+1ln⁡2)g"(n)=\frac{d}{dn}\left(\log_2n+\frac{1}{\ln{2}}\right)
g"(n)=ddnlog⁡2n+0g"(n)=\frac{d}{dn}\log_2n+0
g"(n)=1nln⁡2\boxed{g"(n)=\frac{1}{n\ln{2}}}
t"(n)g"(n)=2(1/nln⁡2)\frac{t"(n)}{g"(n)}=\frac{2}{\left(1/n\ln2\right)}
t"(n)g"(n)=2⋅nln⁡2\frac{t"(n)}{g"(n)}=2\cdot n\ln2
lim⁡n→∞=∞\lim_{n\to\infty}=\infin
∞ implies t(n) has larger order of growth than g(n)\boxed{\infin\text{ implies t(n) has larger order of growth than g(n)}}
t(n)>g(n)t(n)>g\left(n\right)
n2>nlog⁡2nn^2>n\log_2n
let
t(n) =
g(n) =
lim⁡n→∞t(n)g(n)=lim⁡n→∞n23n\lim_{n\to\infty}\frac{t(n)}{g(n)}=\lim_{n\to\infty}\frac{n^2}{3^{n}}

This is an form, so apply L'Hopital's Rule

Apply L hopital Rule

differentiate the numerator

t′(n)=ddn⋅n2=2nt^{\prime}(n)=\frac{d}{dn}\cdot n^2=2n

Differentiate the Denominator

g′(n)=ddn⋅3ng^{\prime}(n)=\frac{d}{dn}\cdot3^{n}
g′(n)=3nln⁡3g^{\prime}(n)=3^{n}\ln3
t′(n)g′(n)=2n3nln⁡3\frac{t'(n)}{g'(n)}=\frac{2n}{3^{n}\ln{3}}
lim⁡n→∞2n3nln⁡3=∞∞\lim_{n\to\infty}\quad\frac{2n}{3^{n}\ln{3}}=\frac{\infty}{\infty}

Apply L hopital Rule, Again

t"(n)=ddn⋅2n=2t"(n)=\frac{d}{dn}\cdot2n=2
g"(n)=ddn3nln⁡3g"(n)=\frac{d}{dn}3^{n}\ln3
g"(n)=ln⁡3ddn⋅3ng"(n)=\ln{3}\frac{d}{dn}\cdot3^{n}
g"(n)=ln⁡3⋅3n⋅ln⁡3g"(n)=\ln{3}\cdot3^{n}\cdot\ln3
g"(n)=(ln⁡3)2⋅3ng"(n)=(\ln{3})^2\cdot3^{n}
t"(n)g"(n)=2(ln⁡3)2⋅3n\frac{t"(n)}{g"(n)}=\frac{2}{(\ln{3})^2\cdot3^{n}}
lim⁡n→∞=2∞\lim_{n\to\infty}=\frac{2}{\infin}
lim⁡n→∞=0\lim_{n\to\infty}=0
0 implies t(n) has smaller order of growth than g(n)\boxed{\text{0 implies t(n) has smaller order of growth than g(n)}}
t(n)<g(n)t(n)<g\left(n\right)
n2<3nn^2<3^{n}

Result

FunctionOrder of Growth
Highest
Highest
Smallest