$$\lim_{x\to\infty}\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}$$
I only managed to show that:
$$8=\left(\frac{2^{\frac{1}{x}}+2^{\frac{1}{x}}}{2}\right)^{3x}\leq\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}\leq\left(\frac{3^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}=27$$
I don't know how to bound it better, or maybe there is so simpler way?
You can apply L'Hospital's rule to limits of this kind; you just have to be a bit clever and introduce a logarithm:
$$\begin{align} \lim_{x\to\infty}\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x} &= \lim_{x\to\infty}\exp\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}\\ &= \exp\lim_{x\to\infty}\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}\\ &= \exp\lim_{x\to\infty}\left(3x\cdot\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)\right)\\ &= \exp\lim_{x\to\infty}\frac{3\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)}{x^{-1}} \end{align}$$
That last limit is of the form $\frac00$, so it's ready for an application of L'Hospital's rule. Can you take it from there?
Hint
$$2^{\frac {1}{x}}=e^{\frac {1}{x}\ln 2}$$
$$=1+\frac {\ln 2}{x}(1+\epsilon_2 (x)) $$
by the same for $3$, the function becomes
$$e^{3x\ln (1+\frac {\ln 6}{2x}(1+\epsilon (x))}$$
using fact that $$\ln (1+X)\sim X \;\;(X\to 0) ,$$
we get the limit $$ 6^{\frac {3}{2}} $$
As it is who the power infinity form we subtract the (1/×)=t
=e^(lim t tends zero [(2^t+3^t-2)/2])×(3/t)
Lim x tends zero( e^x -1)/x =ln(e ) .so (e) replace in above question with 2 and 3 . so you get= e^[ 3/2 (ln2 +ln3 )] = e^[ln (6)^3/2]
=6^(3/2)
Compute first the limit of the (natural) logarithm and do the substitution $x=1/t$: $$ \lim_{t\to0^+}\log\left(\left(\frac{2^t+3^t}{2}\right)^{1/t}\right)= \lim_{t\to0^+}\frac{\log(2^t+3^t)-\log2}{t} $$ This is the derivative at $0$ of $f(t)=\log(2^t+3^t)$; since $$ f'(t)=\frac{2^t\log 2+3^t\log 3}{2^t+3^t} $$ the limit is $\frac{\log 2+\log3}{2}=\log\sqrt{6}$.
Hence the original limit is $\exp(\log\sqrt{6})=\sqrt{6}$.
