evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{\lfloor x^2+x \rfloor }-x}\right) = , $where $\lfloor x \rfloor = $ floor function of $x$
$\bf{My\; Try}::$ for (1) one :: We can write as $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor}$
and we can say that when $x\rightarrow \infty$, Then $\lfloor x\rfloor  \rightarrow x$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln(x)-x}{x} = n\lim_{x\rightarrow \infty}\frac{\ln (x)}{x}-1$
Now Let $\displaystyle L = \lim_{x\rightarrow \infty}\frac{\ln(x)}{x}{\Rightarrow}_{L.H.R} =\lim_{x\rightarrow \infty}\frac{1}{x} = 0$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor} = n\cdot 0-1 =-1$
$\bf{My\; Try}::$ for (2)nd one::we can say that when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
So $\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{x^2+x}-x}\right) = \lim_{x\rightarrow \infty}\frac{\left({\sqrt{x^2+x}-x}\right)\cdot \left({\sqrt{x^2+x}+x}\right)}{\left({\sqrt{x^2+x}+x}\right)}$
$\displaystyle \lim_{x\rightarrow \infty}\frac{x}{\left(\sqrt{x^2+x}+x\right)} = \frac{1}{2}$
Now my doubt is can we write when $x\rightarrow \infty$, Then $\lfloor x\rfloor  \rightarrow x$
and when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
please clear me
Thanks

No. You cannot write so. In this case, going back to the definition is the best. So, in order to solove your questions, use the followings :
$$x-1\lt \lfloor x\rfloor \le x\Rightarrow \frac{x-1}{x}\le \frac{\lfloor x\rfloor}{x}\lt \frac{x}{x}\Rightarrow \lim_{x\to\infty}\frac{\lfloor x\rfloor}{x}=1$$
where $x\gt0.$
