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I need proof the $$\lim_{x\rightarrow 0} (x^3+1)=1$$ symbolically using $\delta$ in terms of $\epsilon$. I think I understand the basic concept of this, however I am having a difficult time when I get to $|x^3|<\epsilon$.

I'm just not sure what to do here. I think I want to separate it into $|x^2||x|<\epsilon$, but I'm not sure exactly why. I think I am having trouble understanding the reason for delta equaling the minimum of two things.

Jonas Meyer
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watcher47
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2 Answers2

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Write the definition. Let $\epsilon > 0$. You want find $\delta > 0$, such that: $$|x - 0| < \delta \implies |(x^3 + 1) - 1| < \epsilon.$$ Put simply, what $\delta > 0$ should you pick, so that: $$|x| < \delta \implies |x^3| < \epsilon \ ?$$

In this question I gave the general strategy for dealing with limits involving polinomials only, using $\epsilon$'s and $\delta$'s, it might be instructive for you to read.

Community
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Ivo Terek
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Okay, you want to find a $\delta$ such that $|f(x) - 1| < \epsilon$ whenever $0 < |x| < \delta$, where $f(x) = x^3 + 1$. Let's start by playing with the $\epsilon$ section. We have $$|x^3 + 1 - 1| < \epsilon$$

We want to manipulate this so it looks like the $\delta$ part. Fortunately, we're keen enough to notice that $1 - 1 = 0$, so we have $$|x^3| < \epsilon$$ In other words, we have $$|x^2||x| < \epsilon$$

That $|x^2|$ is what we have to worry about. Let's, for a second, arbitrarily decide that we want $|x| < 1$. If that's the case, we have $|x^2| < |x| < 1$. Since $|x^2| < 1$, then $|x||x^2| < |x|$. Now simply let $\delta = \min (1, \epsilon)$. Why? If $\epsilon < 1$, then we can also use that. The $1$ comes from the arbitrary decision to make sure $|x| < 1$, which is justified, because we're dealing with short distances.

So is this correct? Let's see:

Let $\delta = \min (1, \epsilon)$ (assume $\epsilon < 1$, and it'll still work out otherwise), then we have $$0 < |x| < \epsilon$$

But since $|x| < 1$ no matter what, then $|x^2| < 1$, and if this is the case, then $|x^2||x| < |x|$. So we have $$|x^2||x| = |x^3| = |x^3 + 1 - 1| = |f(x) - 1| < |x| < \epsilon$$

So $$|f(x) - 1| < \epsilon$$

Which is what we wanted to show.

user28375028
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  • I think this helps me understand it a little more, could you explain why it is important to separate x^3 into x^2 and x? – watcher47 Sep 16 '14 at 00:37
  • It's just a visual aid. Now that I think about it, it really isn't necessary. What's important is that by limiting $x$ to be less than $1$, we know that $|x^2||x| < |x| < \epsilon$, so $|x^3| < \epsilon$. – user28375028 Sep 16 '14 at 01:03