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How do I compute the following limit

$$ \lim_{ n\rightarrow \infty }{ { \left (\frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } }{ 4 } \right ) }^{ n } } $$

$a,b,c,d\geq0$

Thanks.

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6  
People usually include the attempts they've made at solving a problem. It increasing the chance of getting help because it's a lot easier to point out what your doing wrong if you include this information. –  Math_Illiterate Jan 12 '13 at 0:15
    
Are $a,b,c,d$ real non-negative numbers? –  Git Gud Jan 12 '13 at 0:25
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2 Answers 2

up vote 9 down vote accepted

Let $n=\dfrac{1}{m}$ and we have:

$$\lim_{m\to 0}\left(\frac{a^m+b^m+c^m+d^m}{4}\right)^{\frac{1}{m}}$$

This is actually the geometric mean of $a,b,c,d$.

Proof: taking the logarithm, we investigate:

$$\lim_{m\to 0}\dfrac{1}{m}\ln\left(\frac{a^m+b^m+c^m+d^m}{4}\right)$$

Use l'Hopital's treating $m$ as a variable:

$$\lim_{m\to 0}\left(\frac{\ln a\cdot a^{m}+\ln b\cdot b^{m}+\ln c\cdot c^{m}+\ln d\cdot d^{m}}{4}\right)\left(\frac{4}{a^{m}+b^{m}+c^{m}+d^{m}}\right)$$

$$=\dfrac{1}{4}(\ln a+\ln b+\ln c+\ln d)=\ln (abcd)^{\frac{1}{4}}$$

Exponentiate and we get the desired result:

$$\lim_{m\to 0}\left(\frac{a^m+b^m+c^m+d^m}{4}\right)^{\frac{1}{m}}=(abcd)^{\frac{1}{4}}$$

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I think that L'hopital's rule can't be applied to sequences, link –  Msegade Jan 12 '13 at 0:44
3  
See here –  L. F. Jan 12 '13 at 0:55
    
Ok, Thanks for helping!. –  Msegade Jan 12 '13 at 1:02
    
Nice solution!! –  Math_Illiterate Jan 12 '13 at 13:16
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Using the elementary limits $\lim_{n\to\infty} (1+1/n)^n=e$ and $\lim_{x\to 0} (a^x-1)/x=\ln a, \space a>0$, we get

$$\lim_{ n\rightarrow \infty }{ e^{ \left (\displaystyle n \frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d }-4 }{ 4 } \right ) } }={ e^{\displaystyle \frac{1}{4}\lim_{ n\rightarrow \infty } \sum_{abcd} (t^{1/n}-1)/(1/n)}}=(abcd)^{\frac{1}{4}}.$$

Q.E.D.

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hmm ... could you elaborate it? –  Santosh Linkha Jan 12 '13 at 12:50
    
@experimentX: I used $\lim_{ n\rightarrow \infty }{ { \left (1+\frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } }{ 4 } -1\right ) }^{ n } }=\lim_{ n\rightarrow \infty }\left({ { \left (1+\frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } -4 }{ 4 } \right ) }^{ \displaystyle\frac { 4}{ \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } -4 } } }\right)^{n\displaystyle\frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } -4 }{ 4 }}$ –  Chris's sis Jan 12 '13 at 13:13
    
very clever ... –  Santosh Linkha Jan 12 '13 at 13:14
    
@experimentX: thank you. I only used those basic limits I mentioned above. –  Chris's sis Jan 12 '13 at 13:16
    
Can you show all the steps please? –  Msegade Jan 13 '13 at 0:30
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