MHB Proving the Limit of x^n as x Approaches a Using Epsilon-Delta Definition

[Guest2]

How does one prove that $$\displaystyle \lim_{x \to a} x^n = a^n$$ directly from the epsilon-delta definition?

 

[Ackbach]

What have you tried so far?

 

[Guest2]

What have you tried so far?

$\displaystyle |x^n-a^n| < \varepsilon \implies \bigg|x-a\bigg|\bigg|\sum_{k=0}^{n-1}x^k a^{n-k-1}\bigg| <\varepsilon \implies |x-a| < \frac{\epsilon}{ \bigg| \displaystyle \sum_{k=0}^{n-1}x^k a^{n-k-1}\bigg|}$.

However, I don't know how to bound the denominator.

 

[Ackbach]

One way to go about this is the approach my father has in his http://mathhelpboards.com/math-notes-49/method-proving-some-non-linear-limits-4149.html.

There are two parts to this method.

The procedure for obtaining $\delta=\delta_{0}$, (which
will work for all $ \varepsilon\ge \varepsilon_{0}$) is summarized as follows:

1. Select an $ \varepsilon_{0}$ first
2. Take $|f(x)-L|< \varepsilon_{0}$, work it algebraically to
   $L- \varepsilon_{0}<f(x)<L+ \varepsilon_{0}$, and finally work it down to $x_{1}<x<x_{2}$,
   where $x_{1}$ and $x_{2}$ will depend on $f(x)$, $L$, and the $ \varepsilon_{0}$ we
   pick.
3. Take $\delta_{0}$ to be the lesser of $a-x_{1}$, and $x_{2}-a$.
4. Then, for $|x-a|<\delta_{0}$, we know that $|f(x)-L|< \varepsilon$, for
   all $ \varepsilon\ge \varepsilon_{0}$.

Then we need to find $\delta$ that will work for all
$\varepsilon<\varepsilon_0$. To summarize
our method:

1. Take the ratio of $\tfrac{\delta_{0}}{ \varepsilon_{0}}$, both of
   which were determined in the previous section.
2. Check to verify that $|g(x)|$ is indeed a minimum at one
   of the two points $a-\delta_{0}$ and $a+\delta_{0}$.
3. If 2. checks out, then set
   
   $\delta=\tfrac{\delta_{0}}{ \varepsilon_{0}}\, \varepsilon$ for all
   $ \varepsilon< \varepsilon_{0}$.

In the final proof, we're going to set
$$\delta=\min\left(\tfrac{\delta_{0}}{ \varepsilon_{0}}\, \varepsilon,\delta_0\right).$$

So, how does this look for what we have here?

 

[Euge]

How does one prove that $$\displaystyle \lim_{x \to a} x^n = a^n$$ directly from the epsilon-delta definition?

Hi Guest,

Are we to assume that $n$ is a positive integer?

 

[Deveno]

If all you are looking to do is bound:

$\sum\limits_{k = 0}^{n-1} x^ka^{n-k-1}$

you can choose $\delta < a$ (assuming $a > 0$, but the case where $a < 0$ is similar)

so that $0 < x < 2a$

Then $\left|\sum\limits_{k = 0}^{n-1} x^ka^{n-k-1}\right| < \left|\sum\limits_{k = 0}^{n-1} 2^ka^{n-1}\right|< 2^{n-1}na^{n-1}$

and the latter is a constant.

It's a *crude* bound, but that shouldn't matter.

For $a < 0$, choose $\delta < |a|$, so that $0 < |x| < |2a|$, and use the triangle inequality on the polynomial sum.

The basis idea is that we are going to require that $|x - a|$ be small anyway, so $x$ is going to be near $a$, so the sum:

$\sum\limits_{k = 0}^{n-1} x^ka^{n-k-1}$ should be bounded by some expression in $|a|$ ( I just chose $\delta < |a|$ to simplify what otherwise would be rather horrendous algebra).

 

[Euge]

Looking back at Guest's initial work, I assume $n$ is a positive integer. Let $a$ be a fixed real number and suppose $0 < |x - a| < 1$. By the triangle inequality, $|x| = |(x - a) + a| \le |x - a| + |a| < 1 + |a|$. Again, by the triangle inequality,

$$\left|\sum_{k = 0}^{n-1} x^ka^{n-k-1}\right| \le \sum_{k = 0}^{n-1} |x^k a^{n-k-1}| = \sum_{k = 0}^{n-1} |x|^k|a|^{n-k-1} < \sum_{k = 0}^{n-1} (1 + |a|)^k (1 + |a|)^{n-k-1} = n(1 + |a|)^{n-1},$$

and thus

$$|x^n - a^n| < n(1 + |a|)^{n-1}|x - a|,$$

which can be made less than a positive number $\epsilon$ by making

$$|x - a| < \frac{\epsilon}{n(1 + |a|)^{n-1}}.$$

Hence, given $\epsilon > 0$, I choose

$$\delta = \min\left\{1,\frac{\epsilon}{n(1 + |a|)^{n-1}}\right\}.$$

For all $x$, $0 < |x - a| < \delta$ implies $0 < |x - a| < 1$ and $0 < |x - a| < \epsilon/[n(1 + |a|)^{n-1}]$, in which case

$$|x^n - a^n| < n(1 + |a|)^{n-1}|x - a| < \epsilon.$$

Since $\epsilon$ and $a$ were arbitrary, $\lim\limits_{x\to a} x^n = a^n$ for all real numbers $a$.

 

[Deveno]

Looking back at Guest's initial work, I assume $n$ is a positive integer. Let $a$ be a fixed real number and suppose $0 < |x - a| < 1$. By the triangle inequality, $|x| = |(x - a) + a| \le |x - a| + |a| < 1 + |a|$. Again, by the triangle inequality,

$$\left|\sum_{k = 0}^{n-1} x^ka^{n-k-1}\right| \le \sum_{k = 0}^{n-1} |x^k a^{n-k-1}| = \sum_{k = 0}^{n-1} |x|^k|a|^{n-k-1} < \sum_{k = 0}^{n-1} (1 + |a|)^k (1 + |a|)^{n-k-1} = n(1 + |a|)^{n-1},$$

and thus

$$|x^n - a^n| < n(1 + |a|)^{n-1}|x - a|,$$

which can be made less than a positive number $\epsilon$ by making

$$|x - a| < \frac{\epsilon}{n(1 + |a|)^{n-1}}.$$

Hence, given $\epsilon > 0$, I choose

$$\delta = \min\left\{1,\frac{\epsilon}{n(1 + |a|)^{n-1}}\right\}.$$

For all $x$, $0 < |x - a| < \delta$ implies $0 < |x - a| < 1$ and $0 < |x - a| < \epsilon/[n(1 + |a|)^{n-1}]$, in which case

$$|x^n - a^n| < n(1 + |a|)^{n-1}|x - a| < \epsilon.$$

Since $\epsilon$ and $a$ were arbitrary, $\lim\limits_{x\to a} x^n = a^n$ for all real numbers $a$.

For the interested reader, Euge's and my approach are much the same-the idea is to replace $|x|$ with something involving $|a|$. His bound is "tighter", at least for those $a$ where $|a| > 1$.

Indeed, through a "crude" lens, both bounds are on the order of $na^{n-1}$ (the *derivative* of $a^n$ with respect to $a$). This is no accident, the derivative is the BEST possible linear approximation (and $x - a$ is a linear factor of $x^n - a^n$).

One advantage to his method, is it is clear the denominator in one of his possible choices for $\delta$ can never be $0$ (since $1 + |a| > 0$ for all $a$). With my method, the case $a = 0$ has to be handled separately (hopefully it should be clear that this limit is easy to estalbish directly from the definition).

In any case, my main point is with problems like these, there's no "one perfect way to skin the cat". All you have to do is exhibit ANY $\delta > 0$ that works, and there's many,many ways you might derive such a $\delta$.