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Here is the problem. If

$$\lim_{x\to-2}x^3 = - 8$$

then find $\delta$ to go with $\varepsilon = 1/5 = 0.2$.
Is $\delta = -2$?

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    You need to find $\delta$ small enough that if $x$ is within $\delta$ of -2, then $x^3$ is within 0.2 of -8—that is, between 7.8 and 8.2. $\delta=-2$ is not nearly small enough, because then $x$ could be anywhere between -4 and 0, and $x^3$ might be very different from -8.2012-09-19
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    HINT: The idea is that you need to find $\delta$ such that if $$\left| -2 - x \right| \leq \delta$$ then $$ \left|-8 - x^3 \right| \leq \frac{1}{5}$$2012-09-19
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    After Deven's hint, I'll recite my comment: Is it possible that $\delta<0$?2012-09-19
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    really? I plugged in x^3 = -7.8 and x^3 = -8.22012-09-19
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    @Dennis In the problem it does not say that δ has to be greater than 0.2012-09-19
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    @dsta yes but my rephrasing of the problem implies that $\delta$ must be greater than $0$. And $\delta$ measures how "close" we are to $-2$, so you wouldn't want $-2.01$ but rather how close $-2.01$ is to $-2$.2012-09-19
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    It's not about the problem. Take a good look at Deven's hint.2012-09-19
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    δ = min{.02, .01} = .01?2012-09-19
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    @dsta sure $.01$ works, it actually gets you a little closer than you need to be2012-09-19
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    delta cannot be negative? According to the definition 0<|x-a|2013-08-10

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