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Studying real analysis on my own. I've come across the following problem.

$$f(x) = x^2 + 2x + 6; \text{ show } \lim_{x\to3}f(x)=21$$

The solution that was given was:

Choose $\epsilon>0$:

$|f(x)-21|=|x^2+2x-15|$ ...(so far so good we just subtracted the L value from the function)

$=|x-3||x+5|$ ...(normally I would just set $<ε$ and solve for $|x-3|$, but can't do that here as I would still have a variable term)

Then the solution says, restrict $\delta<1$, ..(where $\delta$ represents the size of the window we consider as $x$ approaches $3$. Here is the part that is troubling me. I don't understand why we can do this)

The solution then says:

we want $|x-3|<\delta<1$

$$\implies-1<x-3<1$$

$$\implies->7<x+5<9$$

$$\implies|x+5|<9$$

Then it says:

$$|f(x)-21|=|x-3||x+5|<9|x-3|<\epsilon \text{ if } |x-3|<\epsilon/9$$

Take $\delta=\min(1,\epsilon/9)$

$$\text{Then if } |x-3|<\delta\implies|f(x)-2|<\epsilon$$

$$\lim_{x\to3}f(x)=21$$

I don't understand the intuition behind just restricting $\delta<1$. $\delta$ is the window we must be inside to enable us to get arbitrarily close to the limit value. Why do we and how can we restrict ourselves to the case where $\delta<1$? As I don't really understand this point, I don't understand much of the rest. I don't see why we are taking the minimum value either $\delta=\min(1,\epsilon/9)$ Can somebody help?

JC12
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R.S.
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  • Since $x$ is clearly continuous, and continuous functions are closed under addition and multiplication, $f(x)$ is also continuous. So, $\lim_{x\to3}f(x)=f(3)$ – Kenta S Aug 07 '20 at 09:00
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    @KentaS that's begging the question: you're asked to show that $f$ is continuous at $x=3$ and then say it's obvious because $f$ is continuous... You're assuming what you have to show. – Henno Brandsma Aug 07 '20 at 09:01
  • I'm not assuming $f$ is continuous; I'm saying you can show it from $x$ being continuous, which is obvious. – Kenta S Aug 07 '20 at 09:02
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    In definition we have $\exists \delta >0$, that means we want to find such, that makes inequality true. Most easy answer to your question is - found $\delta$ works or not? If it works, then job is done. Nobody restrict you to find other $\delta$ in more easy way. – zkutch Aug 07 '20 at 09:05
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    See this answer. – Paramanand Singh Aug 07 '20 at 09:07
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    @KentaS plus a lot of theorems on adddition and multiplication being continous etc, all of which has not been covered at this point. All he has is the definition of a limit. – Henno Brandsma Aug 07 '20 at 09:09

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