# Limit of $\sqrt[n]{a^n-b^n}$ as $n\to\infty$

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.

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What did you try to "block" with? if a>b, then $a^n - b^n > 0$. – M.B. Mar 20 '11 at 19:17
Note: the accepted solution deals only with a special case of the question asked (and technically, it does not prove that the limit exists), while another solution, much simpler and shorter, proves the full stuff. – Did Jul 24 '11 at 15:18
Nice problem and very simple. – user 1618033 Jun 11 '12 at 16:15

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 $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L=a.$$
@TonyK: er... hmm... well, conceptually, I agreed with your suggestion/hint (though I didn't carry it forward to see what would happen), but I'm fairly confident in my work under the stated tighter assumption that $a>b>1$, and Mathematica agrees with my result under that assumption. – Isaac Mar 20 '11 at 19:36
@TonyK: Also, assuming $x>1$, Mathematica says that $\lim_{n\to\infty}(1-x^n)^{(1/n)}=x$. – Isaac Mar 20 '11 at 19:38
Ah, so it's me that was wrong. I confused $(1-x^n)^{1/n}$ with $(1-nx)^{1/n}$. Sorry! – TonyK Mar 20 '11 at 19:52
@Isaac: assuming x > 1, that's not a limit you want to take: since 1 - x^n < 0, the n-th root is not even defined. In fact, TonyK's result is much more concise; the factor $– Gerben Mar 20 '11 at 20:02 Rearrange$(a^n-b^n)^{1/n}$as$a(1-(b/a)^n)^{1/n}$. Can you do this? - Definitely, are you suggesting a limit that contains e? – user6163 Mar 20 '11 at 19:25 Should I let a(n)=(b/a)^n and use the e limit theorem. It doesnt seem to be working for me. – user6163 Mar 20 '11 at 19:33 i don't familiar with this. Thank you. – user6163 Mar 20 '11 at 19:44 As Isaac pointed out, I had this completely wrong.$(1-x^n)^{1/n}$tends to 1 if$|x| < 1$. – TonyK Mar 20 '11 at 19:59 The simple way of finishing this off is to note that the expression is between$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$. – Hans Lundmark Mar 20 '11 at 20:01 METHOD I Since$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. - Maybe a little more manipulation needed ... applying the limit to one factor but not the other – GEdgar Jun 11 '12 at 16:22 @GEdgar: thanks for your comments. Do you agree with this way? – user 1618033 Jun 11 '12 at 16:28 @GEdgar: wanting to provide with very short proofs, it happens sometimes that my notation fails ... :) – user 1618033 Jun 11 '12 at 16:34 Better!${}{}{}{}$– GEdgar Jun 11 '12 at 17:11 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 $$\lim_{n\to\infty} (a^n-b^n)^{1/n}=a.$$ - 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 $$a^{1 - {1 \over n}}(a - b)^{1 \over n} < \sqrt[n]{a^n - b^n} <n^{1 \over n} a^{1 - {1 \over n}}(a - b)^{1 \over n}$$ 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.