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We have the function $g(x) = x^3+1$

Prove, with the definition of limit only, that the limit $L=9$ is indeed the limit of the function when $x=2$.

I started with: $\lvert g(x) - L \rvert$ : $\lvert x^3+1-9 \rvert = \lvert x^3-8 \rvert = \lvert (x-2)(x^2+2x+4) \rvert < \epsilon $

From the definition of the limit we know: $ \rvert x-2 \lvert < \delta$, so therefore: $\lvert (x-2)(x^2+2x+4) \rvert < \delta (x^2+2x+4) $

How do I proceed from here? do I need to find a condition for $ x^2+2x+4 $ ?

Thanks guys

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HINT: $x^2+2x+4<19$ for $1<x<3$, hence we can take $\delta=\varepsilon/19$ for $\varepsilon<1$.

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  • $\begingroup$ How did you know that $x2+2x+4 < 19 $ for $ 1 < x < 3 $ ? $\endgroup$ – FigureItOut Nov 14 '14 at 20:46
  • $\begingroup$ @user1326293 It is increasing for positive $x$, and value in 3 is equal to 19. $\endgroup$ – Przemysław Scherwentke Nov 14 '14 at 20:48
  • $\begingroup$ This is not a good answer. You should specify that you are assuming a priori that $\delta < 1 $ so that you can have $|x-2| < 1 $. At the end, you take the min of $1$ and $\epsilon/19$. $\endgroup$ – user139708 Nov 14 '14 at 20:49
  • $\begingroup$ @PedroArbizu It is a HINT! And OP needs some approximation from above. And 1/19 seems to be less than 1. $\endgroup$ – Przemysław Scherwentke Nov 14 '14 at 20:54
  • $\begingroup$ @PedroArbizu Big deal. Let's disregard it's much clearer than the other answer and give downvotes! *facepalm* $\endgroup$ – user2345215 Nov 14 '14 at 20:55
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$$|x^2+2x+4|=|x^2-4x+4+6x-12+12|=|(x-2)^2+6(x-2)+12| \leq |\delta^2+6\delta+12|$$

So for small $\delta$ you have (for example) $|x^2+2x+4|<13$

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