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
 A: It is not true that
$\lfloor x \rfloor
\to x
$.
What is true is that
$\frac{\lfloor x \rfloor}{x}
\to 1
$.
All that you need for
(1) is that
$\frac{\ln x}{x}
\to 0
$.
For (2),
note that
$x^2+x 
\le 
\lfloor x^2+x \rfloor
<
x^2+x+1
$.
Therefore
$\lfloor x^2+x \rfloor
= x^2+x+c
$
where
$0 \le c < 1
$.
You can then write
$\begin{align}
\sqrt{\lfloor x^2+x \rfloor}-x
&=\sqrt{x^2+x+c}-x\\
&=(\sqrt{x^2+x+c}-x)\frac{\sqrt{x^2+x+c}+x}{\sqrt{x^2+x+c}+x}\\
&=\frac{x^2+x+c-x^2}{\sqrt{x^2+x+c}+x}\\
&=\frac{x+c}{\sqrt{x^2+x+c}+x}\\
\end{align}
$
and you can show that this $\to \frac12$
as you did in your answer.
To show that
$\frac{x+c}{\sqrt{x^2+x+c}+x}
\to \frac12$,
note that
$x^2
< x^2+x+c
<x^2+x+1
<(x+1)^2
$,
so
$2x
< \sqrt{x^2+x+c}+x
< 2x+1$.
A: $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} =\lim_{x\rightarrow \infty}\frac{lnx^n}{\lfloor x\rfloor}-1=\lim_{x\rightarrow \infty}\frac{nlnx}{\lfloor x\rfloor}-1$
Now $\displaystyle\frac{nlnx}{x} \le\frac{nlnx}{\lfloor x \rfloor}\le \frac{nlnx}{x-1}$. Hence $\displaystyle \lim_{x\rightarrow\infty}\frac{nlnx}{x}-1\le\lim_{x\rightarrow\infty}\frac{nlnx}{\lfloor x \rfloor}-1\le\lim_{x\rightarrow\infty} \frac{nlnx}{x-1}-1$.
Since $\displaystyle\lim_{x\rightarrow\infty}\frac{nlnx}{x}\to0$ and so does $\displaystyle\lim_{x\rightarrow\infty}\frac{nlnx}{x-1}$. So the required limit $-1$
A: Since $x-1\leq \lfloor x\rfloor \leq x$ we have $$ \frac{n\log{x}}{x}\leq \frac{n\log{x}}{\lfloor x\rfloor}\leq \frac{n\log{x}}{x-1}$$ for $x>1$.  If the left and right converge to zero as $x\rightarrow \infty$ then the center does by the squeeze lemma.  However, the left and right are of the form $\frac{\infty}{\infty}$ so by L'Hospitals rule they converge to zero. Your limit is the middle part minus $1$ as $x\rightarrow \infty$ which is minus $1$.  
A: 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 $x\gt0.$
