# verifing a limit with definition; finding delta when the interval is not circular ...

I want to verify this limit using the definition: $$\lim\limits_{x\to 2}({\frac{2x}{x-1}})=4$$ so, I verify: $$\lim\limits_{x\to 2}(\frac{2x}{x-1})=4 \iff \forall\epsilon>0,\ \exists \delta>0 \text{ s.t. } 0<|x-2|<\delta \implies |\frac{2x}{x-1}-4|<\epsilon$$ I resolve the inequation system: $$\begin{cases} \frac{2x}{x-1}-4<\epsilon\\ \frac{2x}{x-1}-4>-\epsilon \end{cases}$$

obtaining the solution set: $$\frac{4+\epsilon}{2+\epsilon}<x<\frac{4-\epsilon}{2-\epsilon}$$

now for searching $\delta$ $$\frac{4+\epsilon}{2+\epsilon}-2<x-2<\frac{4-\epsilon}{2-\epsilon}-2\iff -\frac{\epsilon}{2+\epsilon} < x-2 <+\frac{\epsilon}{2-\epsilon}$$ but...how can I set $\delta$? the quantities $\frac{\epsilon}{2+\epsilon}$ and $\frac{\epsilon}{2-\epsilon}$ are not equal, so how can I say that $\exists \delta$ s.t. $|x-2|<\delta?$

• Instead of this, it's better to assume a maximum neighborhood like $1$ and find another neighborhood from conditions. Apr 11, 2017 at 14:26
• If $f$ is continuous on an open set containing $a$, then $$\lim_{x\rightarrow a}f(x)=f(a).$$ To see why, note that by continuity, for each $\epsilon>0$, there exists a $\delta>0$ such that for all $x$ satisfying $|x-a|<\delta$, $|f(x)-f(a)|<\epsilon$. But this is just the definition of a limit. Apr 11, 2017 at 14:26

You take the smallest one in absolute value : $\frac{1}{2+\epsilon}<\frac{1}{2-\epsilon}$, so your delta is $\frac{\epsilon}{2+\epsilon}$.
What you need to concern is $\frac{1}{|x - 1|}$ smaller then what. Assume $|x - 2| < \delta$, then \begin{align} |x - 1| = |x - 2 - (-1)| \geq 1 - |x - 2|. \end{align} Hence $1 - |x - 1| \leq |x - 2| < \delta$ and we have $$\frac{1}{|x - 1|} < \frac{1}{1 - \delta}$$ and $$\left| \frac{2x}{x - 1} - 4 \right| = 2\frac{|x - 2|}{|x - 1|} < 2\frac{\delta}{1 - \delta}.$$ Notice that $\delta$ should be less then $1$ or $1 - \delta < 0$. Thus, we assume $$2 \frac{\delta}{1 - \delta} = \epsilon$$ and solve for $\delta$, which is $\frac{\epsilon}{2 + \epsilon} < 1$.