Evaluate $$\lim_{n \to \infty} \sqrt{n^2+2n}-\left[\sqrt{n^2+2n}\right]$$
where $[.]$ is Greatest integer function
My try:
We have $$L=\lim_{n \to \infty} \left\{\sqrt{n^2+2n} \right\}$$ where $\left\{.\right\}$ denotes Fractional part of $x$
we have $$L=\lim_{n \to \infty} \left\{n \left(\sqrt{1+\frac{2}{n}}\right)\right\}$$
Now we have $$\left(\sqrt{1+\frac{2}{n}}\right)=1+\frac{1}{n}-\frac{1}{2n^2}$$
Hence
$$L=\lim_{n \to \infty} \left\{n+\frac{n}{n}-\frac{1}{2n}\right\}$$ So
$$L=\lim_{n \to \infty} 1-\frac{1}{2n}=1$$
Is this right approach?
$$n^2<n^2+2n<n^2+2n+1$$
$$n<\sqrt{n^2+2n}<(n+1)$$
So $\lfloor \sqrt{n^2+2n}\rfloor =n$
So we have to calculate $$\lim_{n\rightarrow \infty}\bigg(\sqrt{n^2+2n}-n\bigg)$$