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

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!

• How do you have $\epsilon<\frac{\epsilon(x-2)}{-2}<\epsilon$??? Commented Apr 26, 2021 at 23:56
• Could you should how delta? Sorry, I'm not too familiar with this. Commented Apr 27, 2021 at 0:07
• Obviously $\epsilon < \frac {\epsilon(x-2)}{-2} < \epsilon$ would mean $\epsilon < \epsilon$ which is obviously false. The only reason we'd have $\epsilon < \frac {\epsilon(x-2)}2$ would be if $\frac {x-2}{2} > 1$. And the only reason we'd have $\frac {\epsilon(x-2)}2 < \epsilon$ would be if $\frac {x-2}2 < 1$. Now I have utterly no idea why you'd think either of these but it clearly can't be both. Commented Apr 27, 2021 at 0:55

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

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.

• Was writing the exact same thing. The $\delta$ restriction and then the $\min$ trick is how you prove epsilon-delta limits in cases like OP. Commented Apr 27, 2021 at 0:24
• @Snoop Its definitely not the most obvious thing to do if your unfamiliar with epsilon-delta proofs Commented Apr 27, 2021 at 0:29

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?

Given $$\epsilon>0$$. To find $$\delta$$, we must add a condition such $$|x-3|<\color{red}{\frac 12} \iff -\frac 12 $$\iff \frac 52 $$\iff \frac 12 $$\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 $$\delta=\min(\color{red}{\frac 12},\frac{\epsilon}{12})$$