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Use the definition of limit to prove that $\lim_{x\to \infty}\frac{x+\lfloor x\rfloor}{x^2}=0$.

Attempt:

to show

$0 < |x - a| < \delta \quad \Rightarrow \quad |f(x) - 0| < \epsilon$, but here $a=\infty$.

Please help me to solve the problem.

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  • $\begingroup$ You need to show that for every $\epsilon$ there exists a $K \geq 0$ such that, if $|x| > K$, $|f(x)| < \epsilon$. This is the definition for "$a = \infty$". $\endgroup$ – Klaramun Mar 12 '17 at 12:30
  • $\begingroup$ You should look at the definition of $$\lim_{x\to\infty}f(x)=L$$ where $L\in\Bbb R$. This is called limit at infinity. $\endgroup$ – Juniven Mar 12 '17 at 12:37
  • $\begingroup$ BTW, what do you mean by $[x]$? Does it refers to the greatest integer function? $\endgroup$ – Juniven Mar 12 '17 at 12:46
  • $\begingroup$ yes, greatest integer function $\endgroup$ – user1942348 Mar 12 '17 at 12:50
  • $\begingroup$ If $\lim_{x\to\infty}f(x)$ be exist then $\lim_{x\to\infty}f(x)=\lim_{n\to\infty}f(n)$. $\endgroup$ – Nosrati Mar 12 '17 at 12:51
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Maybe you need the following definition

Definition We say that $\lim_{x\to\infty}f(x)=L$, where $L\in\Bbb R$, if for every $\epsilon>0$, we can find $K>0$ such that for all $x>K$, we have $|f(x)-L|<\epsilon$.

We have to prove that $$\lim_{x\to \infty}\frac{x+[x]}{x^2}=0,$$ where $[x]$ is called the greatest integer function.


Proof: First, we need to observe that $\forall x\in\Bbb R$, $$x-1<\color\red{[x]\leq x}.$$ So, adding both sides of the inequality $[x]\leq x$ by $x$, we get $$x+[x]\leq 2x.$$ Let $\epsilon>0$. By using the Archimedean Property, we can choose $K\in\Bbb N$ such that $\frac{1}{K}<\frac{\epsilon}{2}$. Thus, if $x>K>0$ then $$\begin{align} \bigg|\frac{x+[x]}{x^2}-0\bigg|&=\bigg|\frac{x+[x]}{x^2}\bigg|\\ &=\frac{|x+[x]|}{|x^2|}\\ &=\frac{x+[x]}{x^2}\\ &\leq \frac{2x}{x^2}=\frac{2}{x}<\frac{2}{K}<\epsilon. \end{align}$$ Apply the definition and we are done.

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I've always been uncomfortable with the epsilonic proofs as they are long and seemingly "tricky" and "avoidable" as I always can use the very intuitive, and less rigorous, idea behind the symbol $\lim$. But it's a prejudice. In fact, the epsilonics use the very same "tricks" that the "usual" procedures. I've enjoyed the try. I think it's correct, but maybe (I am never sure) some points need a more detailed explanation. Anyway, you can point the errors, or the flaws:(

Let be $\epsilon>0$ given, then, let's choose $K=2/\epsilon$. We have

$$\frac{K+[K]}{K^2}=\frac{2/\epsilon+[2/\epsilon]}{(2/\epsilon)^2}\le\frac{2/\epsilon+2/\epsilon}{(2/\epsilon)^2}=\frac{4/\epsilon}{4/\epsilon^2}=\epsilon$$

Consider $x>K$, then

$$\left|\frac{x+[x]}{x^2}\right|=\frac{x+[x]}{x^2}=1/x+(1/x)([x]/x)$$

$$=(1/x)(1+[x]/x)<(1/K)(1+[K]/K)=\frac{K+[K]}{K^2}\le\epsilon\tag 1$$

So, $0$ is the limit.

$(1)$ $x>K>0\implies(1/x<1/K);\;[x]/x\le[K]/K;\;(1+[x]/x)<(1+[K]/K)$ and the inequality follows.

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