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Q) Prove that $ \displaystyle \lim_{x \to 2} x^3 = 8$.

This is how I approached it:

$ \implies |x^3 - 2^3|<\epsilon$

$ \implies |(x-2)(x+2)^2|< \epsilon$

$ \implies |x-2||x+2|^2<\epsilon.$

then I tried to find a bound of $|x+2|^2$,by restricting the x at most away from $1$ i.e $|x-2|<1$;

$-1<x-2<-1$

$3<x+2<5$

$3<|x+2|<5$

$9<|x+2|^2<25$, so $|x+2|^2$ was bounded above by $25$, so,

$|x-2| < \frac{\epsilon}{25}$,

so $\delta$ = min{$1, \frac{\epsilon}{25}$}

Let $\delta$ be $1$, so $1 < \frac{\epsilon}{25}$, $25 < \epsilon$, |x-2| < 1 ==> $|x-2| < 1 < \frac{\epsilon}{25}$

==> $|x-2| < \frac{\epsilon}{25}$

==> $|x-2||x+2|^2 < \frac{\epsilon}{25} * |x+2|^2$

==> $|(x-2)(x+2)^2 | < \frac{\epsilon}{25} *25 $

==> $|x^3 - 2^3| < \epsilon$

==> $|f(x) - L| < \epsilon.$

Is this correct? Do correct me if I was wrong at any point.

Afntu
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Luckyian
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    I think your factorization of $x^3-8$ is wrong. $x^3-8 = (x-2)(x^2+2x+4)$ –  Jan 10 '24 at 07:31
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    The main arguments are OK, you just need to re-arrange it a bit better. After choosing $\delta$, you shouldn't start from the "end" of the previous computation and go backwards, it is less clear in my opinion. A more "clean" approach is to simply repeat the computation, where you write in each paragraph the condition on $\delta$ you used. For example,

    $$ |f(x)-L| = |x^3-8| = |(x-2)(x^2+2x+4)| \le |x-2| |x^2+2x+4| \le _{(1)} |x-2| (9+6+4) = 19 |x-2| < _{(2)} 19 \delta < _{(3)} \varepsilon $$ where (1) follows from $\delta<1$, (2) from $|x-2|<\delta$ and (3) from $\delta < \varepsilon / 19$.

     – Amit Jan 10 '24 at 07:33
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    Why are you letting $\delta = 1$? That only works if $\epsilon > 25$ and proving something only for $\epsilon > 25$ rather than for ALL $\epsilon$ is not useful. – fleablood Jan 10 '24 at 07:34
  • Isn't $|(x +2)|^2 = |x+2|^2$ – Luckyian Jan 10 '24 at 07:49
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    if $\epsilon < 25$, then the minimum value would will be $\frac {\epsilon}{25}$<, then $\delta = \frac {\epsilon}{25}$ right? – Luckyian Jan 10 '24 at 07:51
  • I mean if X is positive then $(x+2)^2=|x+2|^2$, but x is negative then $(-x+2)^2 -> ((-1)x -2)^2 =>(x-2)^2 = |x+2|^2$ – Luckyian Jan 10 '24 at 07:55

1 Answers1

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It seems your factorization of $x^3-8$ is wrong.

$$x^3-8 = (x-2)(x^2+2x+4)$$

Th rest of the procedure is the same as you have done

$$-1 < x-2 < 1$$ Implies

$$1<x<3$$

Or

$$7=1^2+2(1)+4<x^2+2x+4<3^2+2(3)+4=19$$

This means that we can choose $\delta = \min \left(1, \frac{\epsilon}{19} \right)$ such that

$$|f(x)-f(2)| = |x^3 - 2^3| = |x-2| \cdot |x^2+2x+4| \le \delta \cdot (x^2+2x+4) \le \frac{\epsilon}{19} \cdot 19 = \epsilon $$

Which completes the proof.