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I'm a little bit confused by showing the following limit directly from the definition:

$$\lim_{x\to 4} x^3 = 64$$

$|x^3-64|=|x-4| |x^2+4x+16|$

Since I know that $|x-4|< \sigma $, I'm trying to do something like

$|x-4+x^2+3x+20|<\sigma +x^2+3x+20$

Than I multiply two inequalities:

$|x-4||x^2+4x+16|<\sigma^2 +x^2\sigma+3x\sigma+20\sigma$

Here's the point of my confusion. Since we know that x < 4 will always holds, can we multiply right side of inequality to get the following?

$|x-4||x^2+4x+16|<\sigma +16\sigma+12\sigma+20\sigma = 49\sigma$

So then we can take $\sigma = min\{1/2, \sigma/49\}$

Thank you in advance!

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Almost: We know that $4-\sigma<x<4+\sigma$. So we have that $$|x-4||x^2+4x+16|<\sigma |x^2+4x+16|.$$ Since $x^2+4x+16$ has all positive coefficients, $x^2+4x+16$ is increasing when $x>0$. Since $x<4+\sigma$ (and $x>4-\sigma>0$, assume $\sigma < 4$) we have $$\sigma|x^2+4x+16|<\sigma((4+\sigma)^2+4(4+\sigma)+16)$$ and thus $$|x-4|<\sigma\Rightarrow |x^3-64|<\sigma((4+\sigma)^2+4(4+\sigma)+16).$$ Now you can reverse engineer this to make $\sigma((4+\sigma)^2+4(4+\sigma)+16)<\epsilon$.

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In a limit, only the behavior of the function in the neighborhood of the point is important. You can't require that $x < 4$ (that would be taking a side limit!), but you can fix $r$ and work as if the whole dominion was $(4 - r, 4 + r)$, since only the behavior very close to 4 is important (much closer than $r$).

In your case, one could argue something in the lines of

$ |x - 4||x^2 + 4x + 16| < |x - 4||25 + 20 + 16| = 61|x - 4| < 61\delta$

for every $x \in (4 - \delta, 4 + \delta)$, with $\delta$ small enough (in fact, $\delta < 1$, such that $x < 5$).

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