Find limit of $\sqrt{n^3 + 2n} - \sqrt{n^3 + 1}$ as $n$ goes to $+\infty$

Question

Let $a_n = \sqrt{n^3 + 2n} - \sqrt{n^3 + 1}$ . Find $\lim_{n \to \infty} a_n$ if it exists .

My try : Multiplying by the conjugate leads to $$a_n = \frac{2n-1}{\sqrt{n^3 + 2n} + \sqrt{n^3 + 1}} = \frac{2/n - 1/n^2}{\sqrt{1/n +2/n^3} + \sqrt{1/n + 1/n^4}} \ $$ and I've got stuck here . I guess the answer is zero because the first values of $a_n$ have decreased .

Answer 1

You divided by a one power too high. Write it like this: $$\dfrac{2-\frac 1n}{\sqrt{n+\frac 2n}+\sqrt{n+\frac 1{n^2}}}$$

Answer 2

Since $2n-1 \sim 2n$ and $\sqrt{n^3 + 2n} + \sqrt{n^3 + 1}\sim 2n\sqrt{n}$ when $n\to \infty$ the limit is $0$.

$$\lim _{n\to \infty}\frac{2n-1}{\sqrt{n^3 + 2n} + \sqrt{n^3 + 1}}= \lim _{n\to \infty}\frac{2n}{2n\sqrt{n}}=0$$

Answer 3

Hint: Write $$\frac{2n-1}{n^{3/2}\left(\sqrt{1+\frac{2}{n^2}}+\sqrt{1+\frac{1}{n^3}}\right)}$$ and this is $$\frac{n(2-\frac{1}{n})}{n^{3/2}\left(\sqrt{1+\frac{2}{n^2}}+\sqrt{1+\frac{1}{n^3}}\right)}$$

Answer 4

$$\sqrt{n^3+2n}=n^{3/2}(1+2/n^2)^{1/2}=n^{3/2}\left(1+O\left(\frac1{n^2} \right)\right)=n^{3/2}+O(1/\sqrt n).$$ $$\sqrt{n^3+1}=n^{3/2}(1+1/n^3)^{1/2}=n^{3/2}\left(1+O\left(\frac1{n^3} \right)\right)=n^{3/2}+O(n^{-3/2}).$$ Therefore $$\sqrt{n^3+2n}-\sqrt{n^3+1}=O(1/\sqrt n).$$

Answer 5

Short answer:

$$\sqrt{n^3 + 2n} - \sqrt{n^3 + 1}=\frac{2n-1}{\sqrt{n^3 + 2n}+\sqrt{n^3 + 1}}\sim\frac{2n}{2n^{3/2}}\to0.$$