What is limit of $\lim_{x\to\infty}((\frac{a^x+b^x}{2})^{1/x})$?

Question

How I can calculate limite of this equation?! It can be solved using a famous theorem but I forgot it, may someone help me to calculate and prove it or even remind me the theorem?

$$ \lim\limits_{x\to\infty}\left(\frac{a^x+b^x}{2}\right)^{\frac1x} $$

Thanks in advance.

Answer 1

We assume that $a\ge 0$ and $b\ge 0$ and write

$$\begin{align} \lim_{x \to \infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,a\lim_{x \to \infty} \left(\frac{1+(b/a)^x}{2}\right)^{1/x} \end{align}$$

If $a=b$, then the limit is obviously equal to $a$.

Now, we assume without loss of generality that $a>b$. Then, it is easy to see that

$$\begin{align} \lim_{x \to \infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,a\lim_{x \to \infty} \left(\frac{1+(b/a)^x}{2}\right)^{1/x}\\\\ &=a\lim_{x \to \infty} \exp\left(\log \left(\frac{1+(b/a)^x}{2}\right)^{1/x} \right)\\\\ &=a \exp\left(\lim_{x \to \infty} \frac1x \,\log \left(\frac{1+(b/a)^x}{2}\right) \right)\\\\ &=a \exp\left(\lim_{x \to \infty} \left(\frac1x\right) \times \lim_{x \to \infty} \log \left(\frac{1+(b/a)^x}{2}\right) \right)\\\\ &=a \exp\left(0 \times \log \left(\frac12\right) \right)\\\\ &=a \end{align}$$

Of course, one might have observed that inasmuch as (1) $b/a<1$, and (2) $y^{1/x} \to 1$ as $x \to \infty$ for $0<y<1$, then (3) $(b/a)^{x}$ approaches zero as $x \to \infty$ and thus (4) $(\frac{1+(b/a)^x}{2})^{1/x} \to 1$ as $x \to \infty$.

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Note that if the limit were $x \to -\infty$, then we proceed analogously.

We assume that $a\ge 0$ and $b\ge 0$ and write

$$\begin{align} \lim_{x \to -\infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,b\lim_{x \to -\infty} \left(\frac{1+(a/b)^x}{2}\right)^{1/x} \end{align}$$

If $a=b$, then the limit is obviously equal to $b$.

Now, we assume without loss of generality that $a>b$. Then, it is easy to see that as $x \to -\infty$, the term $(a/b)^x$ goes to zero and $(1/2)^{1/x} \to 1$. Thus,

$$\lim_{x \to -\infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} =b$$

Answer 2

HINT: We have $$\dfrac{\max(a^x,b^x)}2 \leq \dfrac{a^x+b^x}2 \leq \max(a^x,b^x)$$ Hence, $$\dfrac{\max(a,b)}{2^{1/x}} \leq \left(\dfrac{a^x+b^x}2\right)^{1/x} \leq \max(a,b)$$

Answer 3

$$\lim_{x\rightarrow \infty }\frac{(a^x+b^x)^{1/x}}{2^{1/x}}$$

now we have two cases as follows if $a\geq b$ $$\lim_{x\rightarrow \infty }\frac{(a^x(1+(b/a)^x)^{1/x}}{2^{1/x}})=\lim_{x\rightarrow \infty }\frac{a(1+(b/a)^{x})^{1/x}}{2^{1/x}}=a$$

and when b>a the limit becomes $b$

Answer 4

Suppose $\;a\ge b>0\;$ , then

$$L:=(a^x+b^x)^{1/x}=b\left(1+\left(\frac ab\right)^x\right)^{1/x}$$

Now observe that

$$\lim_{x\to\infty}\frac{\log\left(1+\left(\frac ab\right)^x\right)}x\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\left(\frac ab\right)^x\log\frac ab}{1+\left(\frac ab\right)^x}\xrightarrow[x\to\infty]{}\log\frac ab$$

so that

$$\left(1+\left(\frac ab\right)^x\right)^{1/x}\xrightarrow[x\to\infty]{}e^ {\log\frac ab}=\frac ab\implies \lim_{x\to\infty}L=b\frac ab=a $$

Of course, $\;2^{1/x}\xrightarrow[x\to\infty]{}1\;$