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I want to compute this limit: $$\lim_{x \to 2} (x-2)= 0$$ by the basic definition of function limit that includes $ \epsilon ,\delta$.as I understood I have to start with $\epsilon$ so: $$ |f(x)-L|<\epsilon \implies |x-2-0| <\epsilon$$ Now goes the part with $\delta$: $$ |x -a| < \delta \implies |x-2|<\delta$$ then, if the expression smaller than $\delta$ equals to the expression smaller than $\epsilon$ shall I assume that in the case $\epsilon = \delta$ ?

and that's the end of the proof?

I'll be happy to know if i did it right and if not, what should i change/do? thanks in advance :)

2 Answers 2

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Because whoever wrote your problem was evil enough to both take the limit as $x \to 2$ and let you evaluate the limit of the expression $x-2$, then yes, that's more or less it. Had any detail of that been different, it might've been easier to keep the different expressions apart, and it would've been a lot easier to see the inner workings of the $\epsilon$-$\delta$ proof (which would be the point of doing it on such an easy example in the first place). But as it stands, you want to pick your $\delta$ such that for the given $\epsilon$, whenever $|x-2|<\delta$ we have $|x-2|<\epsilon$. Setting $\delta = \epsilon$ works just fine.

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    Another sub question, in proofs of this kind i always have to express delta with epsilon, right? No numeral values. – 2017-01-14
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    @Ozk Most of the time, yes. The only real exception is if you want to calculate $\lim_{x \to 5} 3$. In that case, setting $\delta = 1$ is good enough. Also, note that it's quite common also to use $\delta = \min(1, \text{something with }\epsilon)$. For instance, with parabolas, if you don't, it gets trickier than it has to be to find a $\delta$ that works for all _large_ $\epsilon$. – 2017-01-14
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So we have:

$$\lim_{x \to 2} (x-2)= 0$$

This means, $\forall \epsilon>0, \exists \delta > 0, 0<|x-2|<\delta \to |(x-2)-0|<\epsilon$

So we have, $$|(x-2)-0|<\epsilon$$

$$|(x-2)|<\epsilon$$

Since we have what we want, set $\delta = \epsilon$.

Thus, $$0<|x-2|<\delta$$ $$0<|x-2|<\epsilon$$

$$|x-2|<\epsilon$$

Now we are done.