Limit of a sequence : 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.
 A: 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:
$$\lim_{m\to 0}\left(\frac{a^m+b^m+c^m+d^m}{4}\right)^{\frac{1}{m}}=(abcd)^{\frac{1}{4}}$$
A: 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.
