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I have recently come across a problem concerning the limit of a sequence given as, $$\lim_{n\rightarrow \infty} (a^n + b^n)^\frac{1}{n}$$

Where $0<a<b$. Solving the limit is not a concern. But applying L'Hospital's rule to this limit is one to me.

On rearranging we get, $b \lim_{n\rightarrow \infty} ((\frac{a}{b}) ^n + 1)^\frac{1}{n}$. The solver had then applied the L'Hospital's rule to evaluate the limit. Here, as $n\rightarrow \infty$, the limit approaches the form $1^0$. I have tried to think about this limit approaching an indeterminate form. It does not. From my knowledge of Cauchy's Mean Value theorem, I say that we cannot apply the rule to determinate forms. But it seems to me that we can because it worked for the solver. I think I am not clear on this. Can somebody help?

R004
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    You're right, L'Hôpital's rule doesn't apply here. And the limit is $b$. So what answer does the solver give? – Matthew Leingang Oct 28 '17 at 15:18
  • Note that if $n$ is an integer variable then the L'Hospital's Rule does not apply. Rather one can use Cesaro-Stolz under similar circumstances. For the current limit, neither Cesaro-Stolz nor L'Hospital's Rule is applicable. The answer is arrived at easily as $b$ – Paramanand Singh Oct 28 '17 at 15:25
  • To be more specific the situation is like using L'Hospital's Rule for limit $\lim_{x\to \infty} \dfrac{1}{x}$. – Paramanand Singh Oct 28 '17 at 15:27
  • @ParamanandSingh If L'Hôpital's rule gives an answer when $n$ is considered a continuous variable, the limit when $n$ is an integer variable is the same. (It doesn't work in the other direction, for sure.) – Matthew Leingang Oct 28 '17 at 15:30
  • @MatthewLeingang: agree but when the problem is of sequences I prefer to use the theorems on sequences. Or if one wants to apply L'Hospital's Rule or any other method applicable to continuous variable then it is better to be explicit and mention this. – Paramanand Singh Oct 28 '17 at 15:34
  • After agreeing that $n$ may be any positive real number, not necessarily an integer, we can apply L'Hospital's rule to $$\frac{\log \bigl(1 + \bigl(\frac{a}{b}\bigr)^n\bigr)}{n},$$ see e.g here. It is an utterly silly thing to do since we can directly see that the limit is $0$, but it's valid. – Daniel Fischer Oct 28 '17 at 15:46
  • @MatthewLeingang, the solver got $b$ using the rule( quite surprising ). This limit can be solved effortlessly. Applying the rule was not just necessary in my opinion, it was not applicable. The solution comes from Thomas's calculus 12th ed. solution manual. – R004 Oct 28 '17 at 15:59
  • @ParamanandSingh, in "this case", I imagine the sequence to be moving along the curve $(a^x + b^x)^\frac{1}{x}$. So understanding the limit of this curve will automatically help me understand the limit of the sequence. They should be the same in this case. This is how I always apply L'Hospital's rule while evaluating the limit of a sequence. – R004 Oct 28 '17 at 16:06
  • @DanielFischer: Interesting. I didn't realize that there were cases of non-indeterminate forms where the L'Hôpital trick gives the correct limit. But like you say, it's best avoided. I'm more familiar with students trying it for limits such as $\lim_{x\to 0^+} \frac{x}{\cos x}$ and getting $\infty$. – Matthew Leingang Oct 28 '17 at 16:12
  • I made a silly error. The limit can he evaluated using L'Hospital's rule after taking the natural logarithm on both sides. I overlooked the fact that taking the logarithm has nothing to do with L'Hospital's rule. This was very silly of me. Yes, we can apply the rule after taking the logarithm. But that's not necessary here. – R004 Oct 28 '17 at 16:33

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Why not $$y= (a^n + b^n)^\frac{1}{n}\implies \log(y)=\frac{\log(a^n + b^n)}{n}$$ and then $$\frac{a^n \log (a)+b^n \log (b)}{a^n+b^n}$$ Now, divide everything by $b^n$ since $b>a$ to get the limit of $\log(y)$ and then the limit of $y$.

Claude Leibovici
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  • Ah! I now realize where I went wrong. I totally overlooked the fact that taking the natural log on both sides has nothing to do with L'Hospital's rule. It was a silly mistake. Thank you. – R004 Oct 28 '17 at 16:30
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Note that $\log(1+x)\le x$ for $x>-1$. Therefore we have for $a<b$

$$1\le \left(1+\left(\frac ab\right)^n\right)^{1/n}\le e^{\frac1n \left(\frac ab\right)^n}\tag 1$$

whence applying the squeeze theorem to $(1)$ yields the coveted limit

$$\lim_{n\to \infty}\left(1+\left(\frac ab\right)^n\right)^{1/n}=1$$

And we are done!

Mark Viola
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One can take a different view by an expansion which is $$(1 + x)^{1/n} = 1 + \frac{1}{n} \, \frac{x}{1!} + \frac{1}{n} \, \left( \frac{1}{n} - 1\right) \, \frac{x^{2}}{2!} + \mathcal{O}\left(\frac{1}{n^{3}}\right)$$ for which, $0 < a <b$, $$\left(1 + \left(\frac{a}{b}\right)^{n} \right)^{1/n} = 1 + \frac{1}{n} \, \left(\frac{a}{b}\right)^{n} + \frac{1}{2 \, n} \, \left( \frac{1}{n} - 1\right) \, \left(\frac{a}{b}\right)^{2n} + \mathcal{O}\left(\frac{1}{n^{3}}\right)$$ and \begin{align} \lim_{n \to \infty} (a^n + b^n)^{1/n} &= b \, \lim_{n \to \infty} \left(1 + \left(\frac{a}{b}\right)^{n} \right)^{1/n} = 1. \end{align}

If $a=b$ then $$ 2^{1/n} = 1 + \frac{1}{n} + \frac{1}{2 \, n} \, \left( \frac{1}{n} - 1\right) + \mathcal{O}\left(\frac{1}{n^{3}}\right)$$ which leads to $$\lim_{n \to \infty} 2^{1/n} = 1.$$

Leucippus
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