Prove $$\lim_{x\to5}\frac{x+1}{x-2}=2$$ with the $\delta,\epsilon$ limit proof. Find a $\delta$ for any $\epsilon$ which $|x-5|<\delta$ leads to $|\frac{x+1}{x-2} - 2| < \epsilon$.
I used $x-2 \neq 0$ and simplified $|\frac{x+1}{x-2} - 2| < \epsilon$ to $|\frac{x-5}{x-2}| < \epsilon$. But I don't know how to use $|x-5|<\delta$ now. Is the simplification of the inequation usefull?
On
For any $\,\varepsilon>0\,$ there exists $\,\delta=\dfrac{3\varepsilon}{1+\varepsilon}\in\big(0,3\big)$ such that for all $\,x\in\mathbb R\;\land\;x\ne2\;\land\;|x-5|<\delta\;$ it results that
$|x-2|\geqslant3-|x-5|>3-\delta>0\;\;,$
consequently ,
$\begin{align}\left|\dfrac{x+1}{x-2}-2\right|&=\left|\dfrac{x-5}{x-2}\right|<\dfrac\delta{3-\delta}=\dfrac{\frac{3\varepsilon}{1+\varepsilon}}{3-\frac{3\varepsilon}{1+\varepsilon}}=\\&=\dfrac{3\varepsilon}{3+3\varepsilon-3\varepsilon}=\varepsilon\;.\end{align}$
Hence ,
for any $\,\varepsilon>0\,$ there exists $\,\delta=\dfrac{3\varepsilon}{1+\varepsilon}>0\;$ such that for all $\,x\in\mathbb R\;\land\;x\ne2\;\land\;|x-5|<\delta\;$ it results that $\;\left|\dfrac{x+1}{x-2}-2\right|<\varepsilon\;.$
By definition of limit, it means that
$\lim\limits_{x\to5}\dfrac{x+1}{x-2}=2\;.$
Yes, it is useful to simplify $\displaystyle{\left|\frac{x+1}{x-2}-2\right|<\varepsilon}$ to $\displaystyle{\left|\frac{x-5}{x-2}\right|<\varepsilon}$. The next step is to write this as $$ \frac{|x-5|}{|x-2|}<\varepsilon \, . $$ Let's begin by requiring that $|x-5|<1$, so that we can come up with a lower bound of $|x-2|$. If $|x-5|<1$, then $-1<x-5<1$, and so $2<x-2<4$. Since $2$ and $x-2$ are both positive, we can take reciprocals of the inequality $2<x-2$ while reversing the direction of the inequality sign. This yields $$ \frac{1}{2}>\frac{1}{x-2} \, ; $$ multiplying both sides by $|x-5|$, this becomes $$ \frac{|x-5|}{x-2}<\frac{|x-5|}{2} \, . $$ So if we can force $\displaystyle{\frac{|x-5|}{2}}$ to be smaller than $\varepsilon$, then it will certainly be the case that $\displaystyle{\frac{|x-5|}{x-2}=\frac{|x-5|}{|x-2|}}$ is smaller than $\varepsilon$. This amounts to requiring that $|x-5|<2\varepsilon$. Hence, if $|x-5|<1$ and $|x-5|<2\varepsilon$, then $$ \frac{|x-5|}{x-2}<\varepsilon \, , $$ which is what we wanted. Take $\delta=\min(1,2\varepsilon)$. Notice that I didn't consider what I should pick as my $\delta$ until the very end of the proof.
Response to comment: @Angelo: It's not wrong—it's just that when you come up with an epsilon-delta proof yourself, very often you only pick the $\delta$ at the end. My answer tries to mimic the thought process that I would go through. Of course, you can rewrite it more neatly at the end. If $|x-5|<1$ and $|x-5|<2\varepsilon$, then \begin{align} x-2>2 &\implies\frac{1}{x-2}<\frac{1}{2} \\[5pt] &\implies\frac{|x-5|}{x-2}<\frac{|x-5|}{2}<\frac{2\varepsilon}{2}=\varepsilon \, . \end{align}