Trouble with a sequence limit

Question

I'm having problems with this sequence limit. Note: I am not allowed to use L'Hopital's rule.

$$\lim_{n\to\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n^3+1} - n\sqrt{n}}$$

Thanks in advance ^^.

Answer 1

HINT

In the first place, notice that

\begin{align*} \sqrt{n^{2} + 1} - n = \sqrt{n^{2}+1} - n\times\frac{\sqrt{n^{2}+1}+n}{\sqrt{n^{2}+1}+n} = \frac{1}{\sqrt{n^{2}+1}+n} \end{align*}

Analogously, we have

\begin{align*} \frac{1}{\sqrt{n^{3}+1} - n\sqrt{n}} = \frac{1}{\sqrt{n^{3}+1} - n\sqrt{n}}\times\frac{\sqrt{n^{3}+1} + n\sqrt{n}}{\sqrt{n^{3}+1} + n\sqrt{n}} = \sqrt{n^{3}+1} + n\sqrt{n} \end{align*}

From whence we get

\begin{align*} \frac{\sqrt{n^{2}+1}-n}{\sqrt{n^{3}+1}-n\sqrt{n}} = \frac{\sqrt{n^{3}+1}+n\sqrt{n}}{\sqrt{n^{2}+1}+n} \end{align*}

Can you proceed from here?

Answer 2

Hint:

Set $1/n=h^2$ to find

$$\lim_{h\to0^+}\dfrac{h^3}{h^2}\cdot\dfrac{\sqrt{1+h^4}-1}{\sqrt{1+h^6}-1}$$

Now use $\lim_{y\to0}\dfrac{\sqrt{1+y}-1}y=\dfrac12$ which can be derived using $\sqrt{1+y}-1=x$