Checking epsilon delta limit solution

Hi, i've just started learning the epsilon delta limit, and i've done a question, i know that i got the right answer, but i just want to check if my presentation/solution is correct

I've put the link to the quesion and solution here: https://ibb.co/yPY3mZ5
(edited 1 month ago)
Hi, i've just started learning the epsilon delta limit, and i've done a question, i know that i got the right answer, but i just want to check if my presentation/solution is correct
I've put the link to the quesion and solution here: https://ibb.co/yPY3mZ5

Looks like a few typos, so

last line delta = min ...., not epsilon = min.

You have an unnecessary/repetition line after |x+3|<6

you mix <= and < unnecessarily

Hi, i've just started learning the epsilon delta limit, and i've done a question, i know that i got the right answer, but i just want to check if my presentation/solution is correct
I've put the link to the quesion and solution here: https://ibb.co/yPY3mZ5

Take these comments in conjunction with mqb2766's.

The overall form of what you've written is fine. This is more important than the small mistakes you've made.

The typo (epsilon for delta) and unnecessary line really don't matter.

But when you go from |x-2|<=1 to 1<x<3, I'd consider that a "true" mistake (even though it's inconsequential here). You do need to be careful about dealing correctly with <= v.s. <; occasionally you'll have arguments where confusing them gives you an incorrect result.
Thank you for your responses, i've fixed the <= sign ( going from |x-2|<=1 to 1<x<3) and i've fixed the typo making sure it's delta = min ... not epsilon = min...

As for the unnecessary/repetition after the |x+3|<6, I'd like to keep it as long as it's mathematically correct, only because it's what my lecturer wrote.

I've fixed my answers, could someone take a look at this and see if it's okay now?

Link to question and solution: https://ibb.co/bBmSMKv
Thank you for your responses, i've fixed the <= sign ( going from |x-2|<=1 to 1<x<3) and i've fixed the typo making sure it's delta = min ... not epsilon = min...
As for the unnecessary/repetition after the |x+3|<6, I'd like to keep it as long as it's mathematically correct, only because it's what my lecturer wrote.
I've fixed my answers, could someone take a look at this and see if it's okay now?
Link to question and solution: https://ibb.co/bBmSMKv

Well, if you want a proper critique:

$|x-2| < 1$
$1 < x < 3$
$4

is not a mathematical argument, it's just a dump of equations. You should connect them together - it doesn't need much - I would just use "so" [1], so something like "|x-2| < 1, so 1 < x < 3, so 4 < x + 3 < 6" is better.

If you want to make the "final result" stand out (e.g. |x+3| < 6), you can just say "so finally". [2]

The line "your lecturer uses": $|x+3| \leq |x| + 3 < 6$ only adds confusion. Again, think about how you'd connect it to the rest of your argument - you can't, which is a good indication it's superfluous. [3]

$|x^2+x-6| < 6 |x-2| < 6\delta$. I would write $|x^2+x-6| = |x+3||x-2| < 6 |x-2| < 6\delta$ here. In principle, you can write it here and avoid where you did the factorization higher up, but even if you don't, I'd prefer the repetition at the point where you actually use it. It's only 5 seconds of writing.

Your final value $\delta = \min(\frac{\varepsilon}{6}, 1)$ allows $\delta = 1$, but you've assumed earlier that $\delta < 1$.

[1] I recommend "so" over a symbol like $\implies$; the latter can get you tangled in knots because you will have sometimes have expressions like $x > \alpha \implies y > \beta$ and then if you want to argue from that that $y^2 < \beta^2$ and you write $x > \alpha \implies y > \beta \implies y^2 < \beta^2$ it's not clear what implies what. (OK, in this case it's still fairly clear, but there will be times when it isn't).

[2] I'm not saying you have to use these exact words - my main point is that using a few actual words is often more helpful than using symbols.

[3] I'm not clear what you're actually doing here, are you adapting an argument your lecturer used for another question? Because this line is basically "wrong" (it's not literally mathematically incorrect, but it isn't helpful. It's like someone is solving x^2 -5x+6 = 0 and in the middle is a line "$\frac{d}{dx} x^2-5x+6 = 2x - 5$" - it's *true*, but it still shouldn't be there).
(edited 1 month ago)
If |x-2| < d then |x+3| < 5 + d (here d = delta) hence

|x^2 + x - 6| = |x-2||x+3| < d (5+d)

So you can avoid this delta<1 business entirely just by setting d(5+d) = epsilon and solving for d>0.