Expert Answer to "Find the limit of n2logn(1−clog⁡nn)"

Joaquin Salas

Answered question

2022-04-12

Find the limit of

n^{2} \log^{n} (1 - \frac{c \log n}{n})

riasc31lj

Beginner2022-04-13Added 8 answers

Well

\log (1 - c \frac{\log n}{n}) = - c \frac{\log n}{n} + O (\frac{{(\log n)}^{2}}{n^{2}})

and so

n^{2} {\log (1 - c \frac{\log n}{n})}^{n} = {(- c)}^{n} \frac{{(\log n)}^{n}}{n^{n - 2}} {(1 + O (\frac{\log}{n}))}^{n}

Now

n \log (1 + O (\frac{\log n}{n})) = O (\log n)

so that

{(1 + O (\frac{\log n}{n}))}^{n}

grows as at most a polynomial in n. The

n^{n}

in the denominator will swamp everything else...

zakos2zn1mr

Beginner2022-04-14Added 10 answers

The case

c = 0

is clear, and then for the rest of its values it's enough to consider and calculate

lim_{n \to \infty} n^{2} \log^{n} (1 + \frac{| c | \log n}{n})

. Then we have that:

lim_{n \to \infty} n^{2} \log^{n} (1 + \frac{| c | \log n}{n}) = lim_{n \to \infty} n^{2} {(\frac{| c | \log n}{n})}^{n} {({(\frac{\log (1 + \frac{| c | \log n}{n})}{(\frac{| c | \log n}{n})})}^{\frac{n}{| c | \log n}})}^{| c | \log n}

= lim_{n \to \infty} n^{2} {(\frac{| c | \log n}{n})}^{n} \frac{1}{e^{\frac{| c | \log n}{2}}} = 0

The auxiliary limit i resorted to is

lim_{x \to 0} {({\frac{\ln (1 + x)}{x}})}^{\frac{1}{x}} = \frac{1}{\sqrt{e}}

The proof is complete.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?