# 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 $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$

Thanks

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 $2x < \sqrt{x^2+x+c}+x < 2x+1$.

$\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} =\lim_{x\rightarrow \infty}\frac{lnx^n}{\lfloor x\rfloor}-1=\lim_{x\rightarrow \infty}\frac{nlnx}{\lfloor x\rfloor}-1$

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 $-1$

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 $1$.

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 $x\gt0.$