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Question Number 123574 by shaker last updated on 26/Nov/20

Answered by Olaf last updated on 26/Nov/20

$$q\left(n\right)=\frac{{(\sqrt{n^2+3n}-n)}^n}{\sqrt{2n^2+5}}$$$$\ln q\left(n\right)=n\ln (\sqrt{n^2+3n}-n)-\frac{1}{2}\ln (2n^2+5)$$$$\ln q\left(n\right)=n\ln \left[n(\sqrt{1+\frac{3}{n}}-1)\right]-\frac{1}{2}\ln \left[n^2(2+\frac{5}{n^2})\right]$$$$\ln q\left(n\right)=n\ln n+n\ln (\sqrt{1+\frac{3}{n}}-1)-\frac{1}{2}\ln n^2-\frac{1}{2}\ln (2+\frac{5}{n^2})$$$$\ln q\left(n\right)=n\ln (\sqrt{1+\frac{3}{n}}-1)-\frac{1}{2}\ln (2+\frac{5}{n^2})$$$$\ln q\left(n\right)\underset{\mathrm{\infty }}{\sim }n\ln (1+\frac{3}{2n}-1)-\frac{1}{2}\ln (2+\frac{5}{n^2})$$$$\ln q\left(n\right)\underset{\mathrm{\infty }}{\sim }n\ln \left(\frac{3}{2n}\right)-\frac{1}{2}\ln (2+\frac{5}{n^2})$$$$\underset{n\to \mathrm{\infty }}{\lim ln}q\left(n\right)=+\mathrm{\infty }$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }q\left(n\right)=+\mathrm{\infty }$$

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