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Suppose we want to prove $\lim_{x \to 2} g(x) = 4$ and $g(x) = x^2$

So we need to show for all $\epsilon > 0 \space \exists \space \delta>0$ such that $0<|x-2|<\delta \implies |g(x) - 4| < \epsilon$ to complete this proof.

I am having difficulty in following the author's solution.

He says we can make $|x-2|$ as small as we like (makes sense, since $|x-2|<\delta$ for some positive $\delta$)

But we need an upper bound on $|x+2|$. If this term doesn't have an upper bound, then we don't know how to pick an appropriate $\delta$ so this also makes sense.

Now at this point onwards I am failing to understand what is happening.

Author says the $\delta$ neighbourhood is centred at $2$ (okay, since $2$ is the limit point of the domain of $g$), but that is must have a radius no bigger than $\delta = 1$. Why is this the case?

Further more, we get the upper bound $|x+2| \leq |3+2| = 5$

Where does this last line come from?

If we let $\delta = 1$ (why?), this implies $|x-2| < 1$ $\implies$$2-1 < x < 2+1$

So then if $x<3$ and we seek an upper-bound for $|x+2| $, shouldn't it be $|x+2| < 3+2=5$ as opposed to $|x+2| \leq 5$ ?

elbarto
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1 Answers1

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The limit game involves somebody giving you an epsilon, and then you get to choose your delta to try to satisfy the conditions. So, you get to choose to have $\delta \leq 1.$ And that is what the author has done. No matter what $\epsilon$ is served up, you decide to never pick a $\delta >1.$

You're right that the inequality can be made strict, but I expect that they're just using it justify a $\delta = \epsilon/5$ choice, and in that context it doesn't particularly matter. If $|x+2|< 5$ then it is surely the case that $|x+2|\leq 5.$

user24142
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