Using the definition of limit prove the following Using the definition of limit prove that
$$\lim_{x\to \infty}\frac{2x^2+x+1}{x^2-3x+1}=2$$
$$\lim_{x\to \infty}\frac{x-[x]}{x}=0$$ $[x]$ greatest integer function
Please help me to solve such type of problem using $\epsilon$-$K$ method.
 A: For part $1$, we have:
$$2x^2+x+1=2(x^2-3x+1)+7x-1$$
For part $2$, use:
$$0\le x-\lfloor x\rfloor=\{x\}\lt1$$
To use $\epsilon-\delta$, we then need to prove that given $\epsilon\gt0$, we can always find a $\delta$ such that, if $\frac1x\lt\frac1\delta$ (or equivalently $x\gt\delta$), the given $f(x)$ is $\lt\epsilon$.
In the first case, $f(x)=\big|\dfrac{7x-1}{x^2-3x+1}\big|$. Let $\delta=\frac1\epsilon$.
$f(\delta)=\big|\dfrac{1/\epsilon-1}{1/\epsilon^2-3/\epsilon+1}\big|=\big|\dfrac{\epsilon-\epsilon^2}{1-3\epsilon+\epsilon^2}\big|=\epsilon\big|\dfrac{1-\epsilon}{1-3\epsilon+\epsilon^2}\big|\to\epsilon\to0$ as $\epsilon\to0$.
In the second case, we can use $f(x)=|\frac1x|$ and the Squeeze Theorem, and $\delta=\frac1\epsilon$.
$f(\delta)=|\epsilon|=\epsilon$.
A: We need the following:
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
$$\lim_{x\to \infty}\frac{2x^2+x+1}{x^2-3x+1}=2.$$
Proof: Let $\epsilon>0$. Choose $K\in\Bbb N$ such that $K>\frac{9}{2}$ and $\frac{1}{K}<\frac{\epsilon}{21}$. (Hope you get this.). Observe that
$$\begin{align}
x>\frac{9}{2}&\implies 2x>9\\
&\implies 3x-9>x\\
&\implies x-3>\frac{x}{3}\\
&\implies \frac{x-3}{7}>\frac{x}{21}\\
&\implies \frac{7}{x-3}<\frac{21}{x}.
\end{align}$$
Therefore, if $x>K>\frac{9}{2}$ then we get 
$$\begin{align}
\bigg|\frac{2x^2+x+1}{x^2-3x+1}-2\bigg|&=\bigg|\frac{7x-1}{x^2-3x+1}\bigg|\\
&=\frac{7x-1}{x^2-3x+1}\\
&<\frac{7x}{x^2-3x+1}\\
&<\frac{7x}{x^2-3x}\\
&=\frac{7x}{x(x-3)}\\
&=\frac{7}{x-3}<\frac{21}{x}<\frac{21}{K}<\epsilon.
\end{align}$$
Apply the definition and we are done.
