Dethrone
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I read this somewhere:
Somewhere in the middle, you have $\displaystyle -6<x-3<-4$, which, in my textbook says we can rewrite as $|x-3|<6$. Why does that hold? Doesn't that interval consist of $-6<x-3<6$, so you are actually enlarging the domain? Or is that why we need the notation, $\displaystyle \delta=\min\left(1,\frac{\epsilon}{6}\right)$ to say that even if it does takes on values of $-4 < x < 6$ which were part of the added domain, it must always be less than $1$ too?
For the notation $\delta=\min\left(1,\frac{\epsilon}{6}\right)$ , what does that actually mean? Does it mean that the value of which we are actually restricting $\delta$ should actually be the smaller of the two conditions, whatever it may be?
MarkFL said:I can give you an example I posted as a solution at another forum a couple of years ago:
We are given to prove:
[math]\lim_{x\to-2}\left(x^2-x-3\right)=3[/math]
For any given [math]\epsilon>0[/math] we wish to find a [math]\delta[/math] so that:
[math]\left|x^2-x-3-3\right|=\left|x^2-x-6\right|<\epsilon[/math] whenever [math]0<|x+2|<\delta[/math]
To do this, consider:
[math]\left|x^2-x-6\right|=|x+2||x-3|[/math]
Thus, to make:
[math]|x+2||x-3|<\epsilon[/math]
we need only make:
[math]0<|x+2|<\frac{\epsilon}{|x-3|}[/math]
Now, we want to define [math]\delta=\frac{\epsilon}{|x-3|}[/math], but we cannot because [math]\delta[/math] is supposed to be dependent on [math]\epsilon[/math] only. To get around this, we will replace [math]|x-3|[/math] with a number [math]M[/math] which satisfies:
[math]|x-3|\le M[/math]
So now we may write:
[math]|x+2|<\frac{\epsilon}{M}[/math]
and proceed as before, taking [math]\delta=\frac{\epsilon}{M}[/math]. But, we have a problem, as there is no number [math]M[/math] that satisfies:
[math]|x-3|\le M[/math] for all real numbers $x$. However, we are only interested in those close to [math]x=1[/math]. It doesn't matter how close, we just want to bound [math]|x-3|[/math] by restricting $x$ near 1, and any restriction will do. For example, if we require:
[math]0<|x+2|<1[/math], i.e., $x$ should be less than 1 unit away from 1, or equivalently [math]\delta=1[/math], then we have:
[math]-1<x+2<1[/math]
[math]-6<x-3<-4[/math]
We can now take [math]M=6[/math], so we should let [math]\delta=\frac{\epsilon}{6}[/math]. Remember that we also need:
[math]|x+2|<1[/math] so that we define [math]\delta=\min\left(1,\frac{\epsilon}{6}\right)[/math], then:
[math]0<|x+2|<\delta[/math] implies [math]|x+2|<1[/math] and [math]|x+2|<\frac{\epsilon}{6}[/math]. We can now write the proof.
Let [math]\epsilon>0[/math] and define [math]\delta=\min\left(1,\frac{\epsilon}{6}\right)[/math]. Then if [math]0<|x+2|<\delta[/math], we have:
[math]\left|x^2-x-6\right|=|x+2||x-3|[/math]
[math]\left|x^2-x-6\right|<6|x+2|[/math] since [math]|x+2|<\delta[/math] and [math]\delta\le1[/math]
[math]\left|x^2-x-6\right|<6\left(\frac{\epsilon}{6}\right)=\epsilon[/math] since [math]|x+2|<\delta[/math] and [math]\delta\le\frac{\epsilon}{6}[/math]
Therefore, we have shown that [math]0<|x+2|<\delta[/math] implies [math]\left|x^2-x-6\right|<\epsilon[/math] which shows
[math]\lim_{x\to1}\left(x^2-x-3\right)=3[/math] by definition.
Somewhere in the middle, you have $\displaystyle -6<x-3<-4$, which, in my textbook says we can rewrite as $|x-3|<6$. Why does that hold? Doesn't that interval consist of $-6<x-3<6$, so you are actually enlarging the domain? Or is that why we need the notation, $\displaystyle \delta=\min\left(1,\frac{\epsilon}{6}\right)$ to say that even if it does takes on values of $-4 < x < 6$ which were part of the added domain, it must always be less than $1$ too?
For the notation $\delta=\min\left(1,\frac{\epsilon}{6}\right)$ , what does that actually mean? Does it mean that the value of which we are actually restricting $\delta$ should actually be the smaller of the two conditions, whatever it may be?