Give an $ε$-$δ$ proof to show that $$\displaystyle \lim_{x\to1}(3x − 2) = 1$$
please help me with this question.
Asked
Active
Viewed 86 times
1
-
See here for the technique. – Mhenni Benghorbal Oct 03 '13 at 21:58
2 Answers
3
You have $|(3x-2)-1| = 3 |x-1|$. So, if $|x-1| < \delta$, then $|(3x-2)-1| < 3 \delta$.
So, if you want $|(3x-2)-1| < \epsilon$, then choose $\delta = \frac{1}{3}\epsilon$.
copper.hat
- 172,524
-
-
If you choose $\delta$ to have that value, then the difference between $3x-2$ and $1$ will be less than $\epsilon$. You can choose $\delta = \frac{1}{5} \epsilon$ if you want. You just need to find any $\delta>0$ so that the $\epsilon$ bound holds. – copper.hat Oct 03 '13 at 23:15
0
We need to prove that for any $\varepsilon > 0$ there exists a $\delta > 0$ such that $$0<|x-a|<\delta \implies |(3x-2)-1|<\varepsilon.$$ We start by playing around with the second inequality. A little algebra yields: $$|(3x-2)-1|=|3x-3|=|3(x-1)|=|3|\cdot|x-1|<\varepsilon$$$$|x-1|<\frac{\varepsilon}{3}$$
So in other words, another way of writing $|(3x-2)-1| < \varepsilon$ is $|x-1| < \varepsilon / 3$. Now we are reading to write our proof. For any $\varepsilon > 0$, if we require that $|x-1|<\varepsilon/3$, then it follows that $|3x-3|<\varepsilon$, and therefore $|(3x-2) - 1| < \varepsilon$. Therefore, for any $\varepsilon > 0$ there exists a $\delta > 0$ that satisfies the inequalities the limit definition requires, so we can say that $$\lim_{x \to 1}(3x-2)=1$$