lim-n-cos-xln-k-ln-k-n-k-where-k-n-N-x-R-

Question Number 132321 by Raxreedoroid last updated on 13/Feb/21

\lim_{n \to \infty} \frac{\cos (x \ln k)}{{(\ln k)}^{n} \sqrt{k}} = ?

where k, n \in N, x \in R

Answered by Dwaipayan Shikari last updated on 13/Feb/21

\lim_{n \to \infty} \frac{c o s (l o g (k^{x}))}{{(l o g (k))}^{n} \sqrt{k}} = \frac{k^{x i} + k^{- x i}}{\sqrt{k} {(l o g (k))}^{n}} = \frac{k^{x i - \frac{1}{2}} + k^{x i - \frac{1}{2}}}{{(l o g (k))}^{n}} = Φ

I f k > e Φ = 0

k = e Φ = \frac{c o s (l o g (k^{x}))}{\sqrt{k}}

k < e Φ \to \infty

Commented by Raxreedoroid last updated on 13/Feb/21

sir I think the answer is always 0 except for k = 1

since \cos (x) where x \in R is between [- 1, 1]

and \lim_{x \to \infty} \frac{1}{a \cdot b^{x}} = 0

so in that case

\lim_{n \to \infty} \frac{\cos (\log (k^{x}))}{{(\log (k))}^{n} \sqrt{k}} = \frac{[- 1, 1]}{incredibly large} = 0, k \neq 1

Commented by Raxreedoroid last updated on 13/Feb/21

sorry my mistake

\ln (k) where k > 1 is between (0, \infty)

so your answer was right