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