find the limit of $a_{n}=\sqrt[n]{(n+3)^{n+1}}(\sqrt{n^{2}+3}-n)$

Question

can someone help me find the limit of $a_{n}=\sqrt[n]{(n+3)^{n+1}}(\sqrt{n^{2}+3}-n)$ and explain too?

Answer 1

$$\sqrt[n]{(n+3)^{n+1}}(\sqrt{n^{2}+3}-n)=\frac{3(n+3)^{1+\frac{1}{n}}}{n+\sqrt{n^2+3}}=$$ $$=\frac{3(n+3)\sqrt[n]{n+3}}{n+\sqrt{n^2+3}}=\frac{3(1+\frac{3}{n})}{1+\sqrt{1+\frac{3}{n^2}}}\cdot\sqrt[n]{n+3}\rightarrow\frac{3}{2}$$

Answer 2

First write $$ \sqrt[n]{(n+3)^{n+1}}=(n+3)\sqrt[n]{n+3} $$ Now, $$ \lim_{n\to\infty}\log(\sqrt[n]{n+3})= \lim_{n\to\infty}\frac{\log(n+3)}{n}=0 $$ which implies $\lim_{n\to\infty}\sqrt[n]{n+3}=1$. Thus you just need to find $$ \lim_{n\to\infty}(n+3)(\sqrt{n^2+3}-n) $$ If the limit of the corresponding function exists, then it's the same as the limit of the sequence; you can also apply the substitution $x=1/t$: $$ \lim_{x\to\infty}(x+3)(\sqrt{x^2+3}-x)= \lim_{t\to0^+}\frac{1+3t}{t}\frac{\sqrt{1+3t^2}-1}{t}= \lim_{t\to0^+}\frac{(1+3t)(1+3t^2/2+o(t^2)-1)}{t^2} $$