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I'm stuck at evaluating this limit

$$\lim_{k \to \infty} \left( \frac{2}{a^{1/k}+b^{1/k}} \right)^k, \quad a,\, b>0$$

I tried binomial expansion but didn't seem to work. Can anybody give me a hint?

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3 Answers 3

up vote 0 down vote accepted

Hint: consider $$ k \log \left( \frac{2}{a^{1/k}+b^{1/k}} \right) $$ and notice that $$ \log\left( \frac{2}{a^{1/k}+b^{1/k}} \right) = -\frac{1}{k} \frac{\log a + \log b}{2} + O(\frac{1}{k^2}) $$

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Can you explain a bit more what have you used to get this equation? –  ywx Jun 26 at 11:54
    
Just a Taylor expansion, putting $1/k=x \to 0$. –  Siminore Jun 26 at 12:00

Hint

Without any loss of generality, let us assume that $a=bx$ with $x \gt 1$.So $$ \left( \frac{2}{a^{1/k}+b^{1/k}} \right)^k=\frac{2^k}{b(1+x^{1/k})^k}$$

Can you take from here ?

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We can use a standard result.

Suppose that, $\lim_{x \to a} f(x) =1 , \lim_{x \to a} g(x)= \infty$ then $\lim_{x \to a} (f(x))^{g(x)}= e^q $ where $q = \lim_{x \to a} g(x)(f(x)-1)$

Proof : Write $\lim_{x \to a} (f(x))^{g(x)} = \lim_{x \to a} e^{g(x) \ln(f(x))}$ . Since $\lim_{x \to a} f(x)-1 = 0$,using well known result, $\lim_{x \to a} \ln(f(x))= \lim_{x \to a} f(x)-1$.
So $\lim_{x \to a} e^{g(x) \ln(f(x))} = \lim_{x \to a} e^{g(x)(f(x)-1)}$ .


Using This result in this problem, we have ,

$$ q = \lim_{k \to \infty} k(\frac{1-a^{1/k}+1-b^{1/k}}{a^{1/k}+b^{1/k}}) $$ $$ q = \lim_{k \to \infty} \frac{\frac{1-a^{1/k}}{1/k}+\frac{1-b^{1/k}}{1/k}}{a^{1/k}+b^{1/k}} $$ $$ q = -\frac{-\ln(ab)}{2}$$ So desired limit = $e^q = \frac{1}{\sqrt{ab}}$

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