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How do I show that $$\lim_{n\to\infty} \sqrt[n]{a^n+b^n}=\operatorname{max}(a,b)$$ with $a,b\ge0$.

I tried to do this by dividing it in two cases, when $a=b$ and $a\gt b$.

In the case $a\gt b$ I factored $a^n$ like this: $$\lim_{n\to\infty} \sqrt[n]{a^n \left(1+{b^n\over a^n} \right)} = \lim_{n\to\infty} a\sqrt[n]{1+{b^n\over a^n}} = a\lim_{n\to\infty} \sqrt[n]{1+{b^n\over a^n}}$$

Then I expressed it in exponential way. $$a\lim_{n\to\infty} \left({1+{b^n\over a^n}}\right)^{1/n}$$

Now I need to prove that $\lim_{n\to\infty} \left({1+{b^n\over a^n}}\right)^{1/n}=1$ so the whole limit is equal to a. The problem is that I don't know how to take the limit of $\lim_{n\to\infty} \left({1+{b^n\over a^n}}\right)^{1/n}$ I tried to use the natural logarithm, but it ended up like this: $$\lim_{n\to\infty} \log\left({1+{b^n\over a^n}}\right)^{1/n}=\lim_{n\to\infty} {1\over n} \log\left({1+{b^n\over a^n}}\right)=\lim_{n\to\infty} {1\over n} \lim_{n\to\infty}\log\left({1+{b^n\over a^n}}\right)=0$$ What did I do wrong and how can I do it right?

Bach
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davidllerenav
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2 Answers2

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Assume that $a \geq b$. Else, relabel the numbers $a \leftrightarrow b$. This being understood, we have $a = \max(a,b)$. Then $$a =\sqrt[n]{a^n} \leq \sqrt{a^n+b^n} \leq \sqrt{a^n + a^n} = \sqrt[n]{2} a. $$Now apply $\lim_{n \to +\infty}$ on everything, noting that $\sqrt[n]{2} \to 1$. It follows from the squeeze theorem that $$\lim_{n\to +\infty}\sqrt[n]{a^n+b^n} = a,$$ as wanted.

Ivo Terek
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For your specific question you may proceed as follows using

  • $\lim_{x\to 0}\left(1+x \right)^{\frac{1}{x}} = e$ and
  • Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $\Rightarrow 0< q:= \frac{b}{a} < 1 \Rightarrow q^n \stackrel{n \to \infty}{\longrightarrow} 0$

\begin{eqnarray*} \left({1+{b^n\over a^n}}\right)^{1/n} & = & (1+q^n)^{\frac{1}{n}}\\ & = & \left( \underbrace{(1+q^n)^{\frac{1}{q^n}}}_{\stackrel{n \to \infty}{\longrightarrow}e} \right)^{\underbrace{\frac{q^n}{n}}_{\stackrel{n \to \infty}{\longrightarrow}0}}\\ & \stackrel{n \to \infty}{\longrightarrow} & e^0 = 1 \end{eqnarray*}

trancelocation
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