Use the $\epsilon-\delta$ definition of the limit of a function to prove that $\lim_{x \to3}\frac{x}{x-2}=3$.

Question

Problem: Use the $\epsilon-\delta$ definition of the limit of a function to prove that $\lim_{x \to3}\frac{x}{x-2}=3$.

My Solution/Proof: Let $I$ be an open interval containing $c$ and let $f$ be a function defined on $I$, except possibly at $c$. The limit of $f(x)$, as $x$ approaches $c$, is $L$, denoted by $\lim_{x\to c} f(x)=L$, means that given any $epsilon>0$, there exists $detla>0$ such that for all $x=c$, $$If \:\:|x-c|<\delta,\:\:\:then\:\: |f(x)-L|<\epsilon$$ Here, $\lim_{x\to3}\frac{x}{x-2}=3$ .

Given $\epsilon$, let $\delta\leqslant\epsilon$ . We want to show that when $|x-3|<\delta$, then $|\frac{x}{x-2}-3|<\epsilon$ .

We start with $|x-3|<\delta$:
$|x-3|<\epsilon<\frac{\epsilon(x-2)}{-2}<\epsilon$ $ \:\:\to\:\:$ $|-2(x-3)|<\epsilon(x-2)$ $\:\:$ $\to$ $\:\:$ $|\frac{-2x+6}{x-2}|<\epsilon$ $\:\:$ $\to$ $|\frac{x}{x-2}-3|<\epsilon$.

Thus, $\lim_{x\to3}$ $\frac{x}{x-2}=3$

Can you verify for me that my proof is correct? Thank you!

Answer 1

Your solution is flawed in the choice of $\delta$. I can't see any scratch work showing how you obtained $\delta\leq\varepsilon$, but it is not the right choice, and you have some errors in you implications.

I will show some scratch work on how to choose the correct $\delta$.

Given $\varepsilon>0$, we want to find $\delta>0$ such that if $|x-3|<\delta$, then $$\left|\frac{x}{x-2}-3\right|<\varepsilon $$

Now, we have $$\begin{align*} \left|\frac{x}{x-2}-3\right| &=\left|\frac{x-3(x-2)}{x-2}\right|\\ &=\left|\frac{x-3x+6}{x-2}\right|\\ &=2\left|\frac{x-3}{x-2}\right| \end{align*}$$

This is looking good, so lets pretend for now that we have such a $\delta$ in our hands. Continuing, we need to have $$\begin{align*} \frac{2\delta}{|x-2|}<\varepsilon \end{align*}$$

We need to find some lower bound for $|x-2|$. So, lets just say $|x-3|<1/2$. Then $$-1/2<x-3<1/2\Longrightarrow 0<1/2<x-2<3/2$$

Since $x-2>0$, then $|x-2|=x-2$ and we have found our lower bound, given $|x-3|<1/2$.

So, we have $|x-2|>1/2$ which implies $2>1/|x-2|$. Then, $$\frac{2\delta}{|x-2|}<(2\delta)\cdot 2=4\delta $$

This means, if we choose $\delta=\min\{\varepsilon/4,1/2\}$, then we will have the desired result.

Answer 2

More easy is to start with $\left|\frac{x}{x-2}-3\right |=\left|\frac{6-2x}{x-2}\right |$. Now considering $\delta = \frac{1}{2}$ we have $\frac{1}{2} \lt x-2$, so $\left|\frac{6-2x}{x-2}\right | \lt 4 |x-3|$. Can you finish from here?

Answer 3

Given $ \epsilon>0$. To find $ \delta $, we must add a condition such $$|x-3|<\color{red}{\frac 12} \iff -\frac 12<x-3<\frac 12$$ $$\iff \frac 52<x<\frac 72$$ $$\iff \frac 12<x-2<\frac 32$$ $$\implies \frac{1}{x-2}<2$$ $$\implies |\frac{6(x-3)}{x-2}|\le 12|x-3|$$

So, it is sufficient to find $ \delta>0 $ satisfying

$$|x-3|<\frac 12 \text{ and } |x-3|<\delta \implies $$ $$12|x-3|<\epsilon$$

thus $$\delta=\min(\color{red}{\frac 12},\frac{\epsilon}{12})$$