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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.

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  • $\begingroup$ Thank you @Dr.MV, I want the limits in both situation, when $x\to -\infty$ and $x\to +\infty$ $\endgroup$
    – Soheil
    Apr 12, 2015 at 3:08

4 Answers 4

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

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  • $\begingroup$ Thanks a lot user17762, what theorem you used here?! $\endgroup$
    – Soheil
    Apr 11, 2015 at 20:59
  • $\begingroup$ @S0H31L No, theorem. the only thing used here is $a^x \leq \max(a^x,b^x)$ and $b^x \leq \max(a^x,b^x)$. Further, $\max(a^x,b^x) \leq a^x+b^x$. $\endgroup$
    – Adhvaitha
    Apr 11, 2015 at 21:03
  • $\begingroup$ Ok, Thank you @user17762 again. $\endgroup$
    – Soheil
    Apr 11, 2015 at 21:04
  • $\begingroup$ Another question about the answer, your answer works when $x\to +\infty$ and $x\to -\infty$ or only it is the case when $x\to +\infty$?! how about $x\to -\infty$ ?! $\endgroup$
    – Soheil
    Apr 12, 2015 at 3:13
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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, $\;2^{1/x}\xrightarrow[x\to\infty]{}1\;$

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  • $\begingroup$ Thank you @Timbuc, your answer works when $x\to +\infty$ and $x\to -\infty$ or only it is the case when $x\to +\infty$?! how about $x\to -\infty$ ?! $\endgroup$
    – Soheil
    Apr 12, 2015 at 3:12
  • $\begingroup$ When $\;x\to -\infty\;$ the limit below "Observe that..." , both numerator and denominator converge to zero so this limit is zero and thus the whole thing's limit is $\;b\cdot e^0=b\;$ $\endgroup$
    – Timbuc
    Apr 12, 2015 at 7:13
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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 $x \to \infty$.


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,

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

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    $\begingroup$ I don't understand why this answer received a down vote. $\endgroup$
    – Mark Viola
    Apr 12, 2015 at 3:48
  • $\begingroup$ Thank you Dr., how about $x\to-\infty$? $\endgroup$
    – Soheil
    Apr 12, 2015 at 4:13
  • $\begingroup$ Sure. My pleasure. For $x \to -\infty$, one factors out $\min(a,b)$ and proceeds identically. So, the answer is $\min(a,b)$. $\endgroup$
    – Mark Viola
    Apr 12, 2015 at 4:47
  • $\begingroup$ @S0H31L I added the case for $x \to -\infty$ as requested. Let me know if this helps. I just want to give you the best answer possible. $\endgroup$
    – Mark Viola
    Apr 12, 2015 at 5:01
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$$\lim_{x\rightarrow \infty }\frac{(a^x+b^x)^{1/x}}{2^{1/x}}$$

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 $b$

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  • $\begingroup$ Thank you @Hard, your answer works when $x\to +\infty$ and $x\to -\infty$ or only it is the case when $x\to +\infty$?! how about $x\to -\infty$ ?! $\endgroup$
    – Soheil
    Apr 12, 2015 at 3:11

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