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$\lim_{x \to \infty }\sqrt[x]{a^{x}+b^{x}+c^{x}} $;

$a,b,c\in \mathbb{R}$.

I need to find and prove a limit of this sequence.

I know that for example the limit of $a^{x}$ is $\infty$ for $a>1$. And the limit of $\sqrt[x]{a^{x}+b^{x}+c^{x}}$ should be equal $\sqrt[x]{\lim_{x \to \infty }{a^{x}+b^{x}+c^{x}}}$. Am I right?

But I have no idea what to do next.

Green Banana
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2 Answers2

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I assume $a,b,c > 0$.

For $x> 0$, $$ \sqrt[x]{a^x+b^x+c^x} = \left(a^x+b^x+c^x\right)^{1/x} = e^{\frac{1}{x} \ln(a^x+b^x+c^x) }. $$ By continuity of the exponential, it is sufficient to compute the limit of the exponent, $\frac{1}{x} \ln(a^x+b^x+c^x)$, when $x\to \infty$.

Without loss of generality, suppose $a$ is greater than $b$ and $c$ (the other two cases are symmetric). Then we have $$\begin{align} \frac{1}{x} \ln(a^x+b^x+c^x) &= \frac{1}{x} \ln\left(a^x(1+\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x\right) \\ &= \frac{1}{x} \ln\left(a^x\right)+\frac{1}{x} \ln\left(1+\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x\right)\\ &= \frac{x}{x} \ln a+\frac{1}{x} \ln\left(1+\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x\right) \\ &= \ln a+\frac{1}{x} \ln\left(1+\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x\right) \\ \end{align}$$ Since $b < a$ and $c<a$, we have $1+\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x\xrightarrow[x\to\infty]{} 1+0+0=1$, and therefore $\frac{1}{x} \ln\left(1+\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x\right)\xrightarrow[x\to\infty]{} 0$. We get that overall, $$ \frac{1}{x} \ln(a^x+b^x+c^x) \xrightarrow[x\to\infty]{} \ln a $$ and thus $$ \exp\left( \frac{1}{x} \ln(a^x+b^x+c^x) \right) \xrightarrow[x\to\infty]{} e^{\ln a} = a. $$

Now, this was because of our assumption that $a$ was the maximum of $a,b,c$. In general, the answer will be $$ \exp\left( \frac{1}{x} \ln(a^x+b^x+c^x) \right) \xrightarrow[x\to\infty]{} e^{\ln \max(a,b,c)} = \boxed{\max(a,b,c)}. $$ You can check it easily; you may have to deal separately where the case where $a,b,c$ are not all distinct, so that you may have $a>b=c$ or $a=b=c$ for instance. The proof above can be easily modified and the result will not change.

Clement C.
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2

You can't pull the root outside the limit. And I don't think you can use arbitrary reals; if one of them is negative, even with $x$ running on the the positive integers you'll have big problems to give a meaning to the sequence.

On the other hand, when $x$ is used it is usually understood we're dealing with a function defined on some upper unbounded interval. In this case we need positive bases.

Let $r=\max\{a,b,c\}$. Then you can write $$ \sqrt[x]{a^x+b^x+c^x}= r\left(\frac{a^x}{r^x}+\frac{b^x}{r^x}+\frac{c^x}{r^x}\right)^{1/x} $$ Now, $$ 0<\frac{a^x}{r^x}\le1,\quad 0<\frac{b^x}{r^x}\le1,\quad 0<\frac{c^x}{r^x}\le1 $$ and one of them is $1$. Therefore $$ 1\le \frac{a^x}{r^x}+\frac{b^x}{r^x}+\frac{c^x}{r^x}\le 3 $$ so that $$ 1\le\left(\frac{a^x}{r^x}+\frac{b^x}{r^x}+\frac{c^x}{r^x}\right)^{1/x} \le 3^{1/x} $$ Since $\lim_{x\to\infty}3^{1/x}=1$, the comparison theorem tells you that $$ \lim_{x\to\infty}\left(\frac{a^x}{r^x}+\frac{b^x}{r^x}+\frac{c^x}{r^x}\right)^{1/x}=1 $$

Thus $$ \lim_{x\to\infty}\sqrt[x]{a^x+b^x+c^x}=r=\max\{a,b,c\} $$

egreg
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