Mathematics Asked on February 7, 2021
It’s not allowed to apply L’Hopital’s rule .
I tried using binomial distribution $2^n=(1+1)^n$ in order to manipulate with numerator but don’t know what to do with denominator.
P.s Should i use Squeeze Theorem to this kind of limit .
I need help with this Calculus I problem …
Answer :
$$lim_{nto+infty} frac{2^n}{n^k}=lim_{nto+infty}frac{e^{nln(2)}}{e^{kln(n)}}= lim_{nto+infty}e^{n(ln(2)-kfrac{ln(n)}{n}) } $$ Note that $$ lim _{nto+infty} kfrac{ln(n)} {n} =0$$
So :
$$lim_{nto+infty} frac{2^n}{n^k}=lim_{nto+infty} e^{nln(2)}=+infty $$
Because $ln(2)>0$.
Answered by user839911 on February 7, 2021
Suppose $k>0$. Let $m=[k]+1$. Then, for $n>2m+2$ $$ frac{2^n}{n^k}=frac{(1+1)^n}{n^k}ge frac{binom{n}{m}}{n^{m+1}}=frac{n(n-1)cdots(n-m+1)}{m!n^{m+1}}ge frac{(n-m+1)^{n-m}}{m!n^{m+1}}=frac{1}{m!}(1-frac{m}{n})^{m+1}(n-m)^{n-2m-1} $$ Since $$lim_{nto infty}frac{1}{m!}(1-frac{m}{n})^{m+1}(n-m)^{n-2m-1}=infty $$ hence one has $$ lim_{nto infty}frac{2^n}{n^k}=infty. $$
Answered by xpaul on February 7, 2021
Consider the sequence $a_n=frac{2^n}{n^k}$.
$$frac {a_{n+1}}{a_n}= 2frac{n^k}{(n+1)^k}=2left(frac{n}{n+1} right)^k =2left(1-frac{1}{n+1} right )^k.$$
Since $k$ is fixed, for large $n$ this ratio exceeds, say $frac 32$, so $lim a_n = infty$.
Answered by Robert Shore on February 7, 2021
Take the logarithm of the expression (assuming $k ge 1$)
$$ lnleft(frac{2^n}{n^k}right) =ln(2^n)-ln(n^k) = nln(2) +lnleft(frac{1}{n^k} right) $$ It follows begin{align} lim_{nto infty}lnleft(frac{2^n}{n^k}right) &= lim_{nto infty}left[nln(2) +lnleft(frac{1}{n^k} right)right] \ &=lim_{nto infty}(nln(2)) + lim_{nto infty}lnleft(frac{1}{n^k} right) \ &=+infty + 0 = +infty end{align} So the limit of the logarithm is $+infty$ and so is the limit $$ lim_{nto infty}left(frac{2^n}{n^k}right) = +infty $$ simply explained, $ln(x) to +infty implies x to +infty$
Answered by Physor on February 7, 2021
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