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I am trying to prove this limit to be true: $$\lim_{x\to a}(x^2)=(a^2)$$ using the Epsilon Delta Limit Definition.

So far I can understand how it works but I got stumped on this inequality $$|x+a|<|2a|+1$$ if $|x-a|<1$

I saw this on this following link: https://www.ma.utexas.edu/users/nrauh/teaching/m408d/limits.pdf

I would really appreciate it if someone can explain how is the inequality derived.

Last but not least, a quick question: What does the $$\delta=min\{\frac{\epsilon}{|2a|+1},1\}$$ mean? Is it the range of values that $\delta$ can accept in an open interval with $\frac{\epsilon}{|2a|+1}$ being the minimum value and 1 being the maximum value?

Thank you, appreciate it lots!

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Suppose $|x-a|<1$. This is equivalent to $-1<x-a<1$. If we add $2a$ to this inequality, we get $-1+2a<x+a<1+2a$. Therefore, $$|x+a|<\max\{|-1+2a|,|1+2a|\}\leq 1+|2a|,$$ where for the last inequality, we used the fact that $|-1+2a|\leq |-1|+|2a|=1+|2a|$, and $|1+2a|\leq 1+|2a|$.

As for your second question, $$\delta=\min\{x,y\}$$ means that $\delta$ is chosen to be the minimum of $x$ and $y$, whatever this is. Usually, the reason to do something like this comes about when we want to ensure that two limit "properties" are satisfied at the same time. In your problem, it has to do with needing to have $|x-a|$ small enough while also keeping $|x+a|$ bounded by some constant.

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  • $\begingroup$ Hello there Kevin, thanks for your help, appreciate it. But may I know is the inequality $|x+a|<\max\{|-1+2a|,|1+2a|\}\leq 1+|2a|$ would essentially imply $|x+a|<\|1+2a|\leq 1+|2a|$ since I am choosing either x or y, whichever is smaller? Is it true that if x<y, $\min\{x,y\}$ will be x and vice versa? $\endgroup$ – Derp Apr 20 '15 at 19:34
  • $\begingroup$ Your first question is confusing; there is nothing about a minimum. Your question: "Is the inequality ...(inequality with max)... would essentially imply $|x+a|\leq ... <1+|2a|$ since I am choosing either x or y, whichever is smaller." What do you mean by choosing whichever is smaller? I'm a bit confused here. As to the second question, yes. If $x<y$, then $\min\{x,y\}=x$ and vice versa. $\endgroup$ – Kevin Church Apr 21 '15 at 6:15
  • $\begingroup$ Oh okay I seem to get it now, thank you very much for your help, appreciate it! $\endgroup$ – Derp Apr 23 '15 at 7:31

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