I have this exercise "Use the formal definition of limit to verify the indicated limit."

$ \lim_{x \rightarrow 1} (3x + 1) = 4 $ ...

I have made an attempt but Im kinda stuck.

So I guess it goes something like this;

I have to find $ 0 < |x-1| < \delta \Rightarrow |(3x+1) - 4| < \epsilon$

Just simplyifying to $|3x-3| < \epsilon$

Then I see that I can multiply by both sides in the inequality including delta; $3|x-1| < 3\delta \Leftrightarrow |3x-3| < 3\delta$

And this is where im not sure.. As far as I have understood im supposed to express $\delta$ as a function of $\epsilon$ .. ?

$3\delta = \epsilon \Leftrightarrow \delta = \frac{\epsilon}{3}$

And now im completely stuck, what now? I have no idea

  • $\begingroup$ You're not stuck, you're done with the problem. The objective is to just find a $\delta > 0$, as a function of $\epsilon$, which would make the inequality true. $\endgroup$ – user230734 Aug 27 '15 at 11:02

You've done everything you need! The formal definition of the limit needs you to demonstrate the existence of a $\delta > 0$ for every $\epsilon > 0$ such that $|x-1| < \delta \implies |(3x+1)-4| < \epsilon$. That's exactly what you have shown. In this case, you have shown that whenever we use a $\delta=\frac{\epsilon}{3} > 0$, it does the trick. In other words, for every $\epsilon>0$, we have found a $\delta > 0$ as required by the definition of the limit. We have this formally shown it exists and is equal to $4$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.