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With the initial conditions: $a>b>0$;

I need to find $\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$

I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.

  • 0
    Nice problem and very simple.2012-06-11

5 Answers 5

2

Another way: Note that $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$. Since $b < a$, each term in the sum is bounded by $a^{n-1}$ and we have $a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} < na^{n-1}$ Since the sum is at least the first term, we also have $a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} > a^{n-1}$ Combining we have $a^{n-1}(a - b) < a^n - b^n < na^{n-1}(a - b)$ Taking $n$th roots we get $a^{1 - {1 \over n}}(a - b)^{1 \over n} < \sqrt[n]{a^n - b^n} This can be rewritten as $a \bigg({a - b \over a}\bigg)^{1 \over n} < \sqrt[n]{a^n - b^n} < a n^{1 \over n} \bigg({a - b \over a}\bigg)^{1 \over n}$ As $n$ goes to infinity both the left and right sides of the above go to $a$. Thus by the squeeze theorem so does the middle and we are done.

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    nice proof. btw, for the right side is enough to remove the second term under the radical. :)2012-06-11
11

Rearrange $(a^n-b^n)^{1/n}$ as $a(1-(b/a)^n)^{1/n}$. Can you do this?

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    The simple way of finishing this off is to note that the expression is between $a(1-b/a)^{1/n}$ and $a$. Now use the squeeze theorem and the fact that the $n$th root of a constant tends to 1 as $n \to \infty$.2011-03-20
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METHOD I

Since $a>b>0$, our limit boils down to:

$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=\lim_{n\to\infty}\sqrt[n]{a^n \left(1-({\frac{b}{a})^{n}}\right)} = \lim_{n\to\infty}\sqrt[n]{a^n}\lim_{n\to\infty}\sqrt[n]{1-(\frac{b}{a})^{n}}=a.$

METHOD II

We may simply squeeze it:

$a= \lim_{n\to\infty}\sqrt[n]{(a-b)a^{n-1}}\leq \lim_{n\to\infty}\sqrt[n]{a^n-b^n} \leq \lim_{n\to\infty}\sqrt[n]{a^n}=a$

The proofs are complete.

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    Better!${}{}{}{}$2012-06-11
4

Suppose the limit exists and is $\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L.$ Then, $\begin{align} \log L &=\log\lim_{n\to\infty}\sqrt[n]{a^n-b^n} \\ &=\lim_{n\to\infty}\log\sqrt[n]{a^n-b^n} \\ &=\lim_{n\to\infty}\frac{1}{n}\log(a^n-b^n) \\ &=\lim_{n\to\infty}\frac{\log(a^n-b^n)}{n}. \end{align}$ If $a>b>1$ (note: $1$, not $0$—I am not sure offhand how to handle the case where $b$ and possibly $a$ are less than 1), $a^n-b^n\to\infty$, so $\log(a^n-b^n)\to\infty$, so the limit is of the indeterminate form $\frac{\infty}{\infty}$ and L'Hôpital's rule applies. $\begin{align} \lim_{n\to\infty}\frac{\log(a^n-b^n)}{n} &=\lim_{n\to\infty}\frac{\frac{d}{dn}\log(a^n-b^n)}{\frac{d}{dn}n} \\ &=\lim_{n\to\infty}\frac{\frac{1}{a^n-b^n}\cdot\frac{d}{dn}(a^n-b^n)}{1} \\ &=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot(a^n\log a-b^n\log b) \\ &=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot(a^n\log a-b^n\log a+b^n\log a-b^n\log b) \\ &=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot((a^n-b^n)\log a+b^n(\log a-\log b)) \\ &=\lim_{n\to\infty}\left(\log a+\frac{b^n}{a^n-b^n}(\log a-\log b)\right) \\ &=\log a+(\log a-\log b)\lim_{n\to\infty}\frac{1}{(\frac{a}{b})^n-1} \end{align}$ and since $a>b>1$, $\frac{a}{b}>1$, so $(\frac{a}{b})^n\to\infty$ and $\lim_{n\to\infty}\frac{1}{(\frac{a}{b})^n-1}=0$, so $\log L=\log a$ and $\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L=a.$

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    @Isaac: See my solution for the general problem.2011-03-21
4

Here is a short solution based on standard inequalities.

Our first inequality is obvious since $b^n>0$ $(1)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\leq a^n.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$

Next we note that $a^n-b^n = a\cdot a^{n-1}-b\cdot a^{n-1}+ b\cdot a^{n-1}- b\cdot b^{n-1} =(a-b)a^{n-1} + b(a^{n-1}-b^{n-1})$ which together with $a^{n-1}- b^{n-1}\ge0$ leads to $(2)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\geq (a-b)a^{n-1}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $

Combining (1) with (2) and taking the $n$:th-root we get $ a \ge (a^n-b^n)^{1/n}\ge(a-b)^{1/n}\cdot a^{(n-1)/n}= a\cdot(a-b)^{1/n}\cdot a^{-1/n} $ where the right hand side tends to $a$ as $n\to\infty$, and hence we reach $\lim_{n\to\infty} (a^n-b^n)^{1/n}=a.$