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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

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  • $\begingroup$ 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
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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 $4$.

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