1
$\begingroup$

Here is the problem. If

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

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

  • 0
    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
  • 3
    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
  • 0
    After Deven's hint, I'll recite my comment: Is it possible that $\delta<0$?2012-09-19
  • 0
    really? I plugged in x^3 = -7.8 and x^3 = -8.22012-09-19
  • 0
    @Dennis In the problem it does not say that δ has to be greater than 0.2012-09-19
  • 0
    @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
  • 0
    It's not about the problem. Take a good look at Deven's hint.2012-09-19
  • 0
    δ = min{.02, .01} = .01?2012-09-19
  • 0
    @dsta sure $.01$ works, it actually gets you a little closer than you need to be2012-09-19
  • 0
    delta cannot be negative? According to the definition 0<|x-a|2013-08-10

1 Answers 1

2

Sometimes Calculus students are under the impression that in situations like this there is a unique $\delta$ that works for the given $\epsilon$ and that there is some miracle formula or computation for finding it.

This is not the case. In certain situations there are obvious choices for $\delta$, in certain situations there are not. In any case you are asking for some $\delta\gt 0$ (!!!) such that for all $x$ with $|x-(-2)|\lt\delta$ we have $|x^3-(-8)|\lt 0.2$. Once you have found some $\delta\gt 0$ that does it, every smaller $\delta\gt 0$ will work as well.

This means that you can guess some $\delta$ and check whether it works. In this case this is not so difficult as $x^3$ increases if $x$ increases. So you only have to check what happens if you plug $x=-2-\delta$ and $x=-2+\delta$ into $x^3$ and then for all $x$ with $|x-(-2)|$ you will get values of $x^3$ that fall between these two extremes.

For an educated guess on $\delta$, draw a sketch. This should be enough information to solve this problem.

  • 0
    Am I correct with δ = min{.02, .01} = .01?2012-09-19
  • 0
    Plug $-2.01$ and $-1.99$ into $x^3$ and see what happens.2012-09-19