$\displaystyle Let \ L=\lim_{x\to \infty} \left ( \dfrac {1^\frac {1}{x}+ 2^{\frac {1}{x}}+ 3^{\frac {1}{x}}+ 4^{\frac {1}{x}}} {4} \right )^{4x} $ $plug \ in \ x=\dfrac {1}{y} \ as \ x\to \infty, \ y \to 0$ $\displaystyle L=\lim_{y \to 0} \left ( \dfrac{1^y + 2^y + 3^y +4^y}{4} \right )^{\frac {4}{y}}$ Taking log on both sides, $\displaystyle \log L= \log \lim_{y \to 0} \left ( \dfrac {1^y+2^y+3^y+4^y}{4} \right )^{\frac {4}{y}}$ Since, log is a continuous function, we can write, $\displaystyle \log L=\lim_{y\to 0}\log \left ( \dfrac {1^y+2^y+3^y+4^y}{4} \right )^{\frac {4}{y}} = \lim _{y\to 0} \dfrac {4}{y} \log \left ( \dfrac {1^y+2^y +3^y+4^y}{4} \right )$ $\displaystyle =\lim_{y\to 0} \dfrac {4[\log (1^y+2^y+3^y+4^y)-\log 4]}{y}$ Now, this comes out to be $\dfrac {0}{0}$ form, hence we can use L' Hospital's rule, $\displaystyle = \lim_{y\to 0} \dfrac {4 \frac {d}{dy}[\log (1^y+2^y+3^y+4^y)-\log 4]}{\frac {d}{dy}y}$ $\displaystyle = \lim_{y\to 0}\dfrac {4}{1} \dfrac {1}{(1^y+2^y+3^y+4^y)}\dfrac {d}{dy} (1^y+2^y+3^y+4^y)-0 $ $\displaystyle = \lim_{y \to 0} \dfrac {4 (1^y \log 1 +2^y \log 2 +3^y \log 3 + 4^y \log 4)}{(1^y+2^y+3^y+4^y)}$ Directly plug in y=0, $=\dfrac {4(1^0 \log 1+2^0 \log 2 +3^0 \log 3 +4^0 \log 4)}{(1^0+2^0+3^0+4^0)} = \dfrac {4(\log 1+ \log 2+ \log 3 + \log 4)} {4}$ $=\log (1\times 2 \times 3 \times 4)$ $\therefore \log L= \log 24 \to L=24$ $\therefore \displaystyle \lim_{x \to \infty} \left ( \dfrac {1^{\frac {1}{x}}+ 2^{\frac {1}{x}}+ 3^{\frac {1}{x}}+ 4^{\frac {1}{x}}} {4} \right )^{4x} = 24$