Evaluation of $\lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}$ 
Evaluation of $\displaystyle \lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}$

$\bf{My\; Try::}$ I am Using above question using Sandwich Theorem
So Using $\bf{A.M\geq G.M\;,}$ We get
$$\frac{16^x+9^x}{2}\geq (16^x\cdot 9^x)^{\frac{1}{2}}\Rightarrow \lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}\geq \lim_{x\rightarrow 0}(16^x\cdot 9^x)^{\frac{1}{2x}}=12$$
But I did not Understand How can I Calculate it for Upper bond, Help me
Thanks
 A: With
$$
L = \left( \frac{16^x + 9^x}{2} \right)^{1/x}
$$
we have
\begin{align}
\lim_{x\to 0} \ln(L) 
&= \lim_{x\to 0}\frac{\ln(16^x + 9^x) - \ln(2)}{x} \\
&= \lim_{x\to 0}\frac{\frac{16^x \ln(16) + 9^x \ln(9)}{16^x + 9^x}{}}{1} \\
&= \frac{\ln(16) + \ln(9)}{2} \\
&= \ln(144)/2
\end{align}
where L'Hôspital was used and $(a^x)'= a^x \ln(a)$. This gives
$$
\lim_{x\to \infty} \ln(L) = \ln(\sqrt{144}) = \ln(12) \Rightarrow \\
\lim_{x\to \infty} L  = 12
$$
because the natural logarithm is continous.
A: This is just a small variant on some of the other answers that have taken logs and used L'Hopital's rule to evaluate the limit.
Note that
$${16^x+9^x\over2}=12^x\left((4/3)^x+(3/4)^x\over2 \right)=12^x\cosh\lambda x\quad\text{with}\quad\lambda=\ln(4/3)$$
so it suffices to show that
$$\lim_{x\to0}(\cosh\lambda x)^{1/x}=1$$
which is equivalent to showing
$$\lim_{x\to0}{\ln(\cosh\lambda x)\over x}=0$$
This last limit is really just the limit definition of the derivative $f'(0)$ for $f(x)=\ln(\cosh\lambda x)$.  We have $f'(x)={\lambda\sinh\lambda x\over\cosh\lambda x}$, so $f'(0)=0$, as desired.  We find that
$$\lim_{x\to0}\left(16^x+9^x\over2 \right)^{1/x}=12\lim_{x\to0}(\cosh\lambda x)^{1/x}=12\cdot1=12$$
A: An alternate approach: the logarithm is $$\frac{\ln\mathopen{}\left(16^x+9^x\right)\mathclose{}-\ln(2)}{x}=\frac{\ln\mathopen{}\left(1+(9/16)^x\right)\mathclose{}+x\ln(16)-\ln(2)}{x}=
\ln(16)+\frac{\ln\mathopen{}\left(1+(9/16)^x\right)\mathclose{}-\ln(2)}{x}$$
L'Hospital says the last fraction has limit the same as $$\frac{(9/16)^x\ln(9/16)}{1+(9/16)^x},$$ whose limit is $\frac{\ln(9/16)}{2}$. So the limit of the log of your expression is $$\ln(16)+\frac{\ln(9/16)}{2}=\ln(9\cdot16)/2$$ and so the limit of your expression is $\sqrt{9\cdot16}=12$.
A: $$\lim_{x\to0}\left(\frac{16^{x}+9^{x}}2\right)^{1/x}\\=\lim_{x\to0}\exp\left(\frac 1x\ln \left(\frac{16^x+9^x}2\right)\right)\\\stackrel{L'H}=\lim_{x\to0}\exp\left(\frac2{16^x+9^x}\cdot\frac12\cdot(16^x\ln 16+9^x\ln 9)\right)\\=\exp\left(\frac{2}2\cdot\frac12\cdot\ln (16\cdot 9)\right)=\exp(\ln\sqrt{16.9} )=4.3=12$$
A: looking at a particular sequence $x = \frac1n \rightarrow 0$ and using
$$
\lim_{n \rightarrow \infty} n(\sqrt[n]{a}-1) = \log a
$$
gives 
$$
\sqrt[n]{a} \rightarrow \frac{\log a}{n} +1
$$
so
$$
\frac12(16^{\frac1n}+9^{\frac1n}) \rightarrow 1+\frac1n \log 12
$$
and
$$
(1+\frac1n \log 12)^n \rightarrow \exp(n \times\frac1n \log 12) = 12
$$
A: Whenever the base and the exponent are variables it makes sense to take logs. If $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{16^{x} + 9^{x}}{2}\right)^{1/x}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{16^{x} + 9^{x}}{2}\right)^{1/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(\frac{16^{x} + 9^{x}}{2}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\cdot\frac{16^{x} + 9^{x} - 2}{2}\cdot\dfrac{\log\left(1 + \dfrac{16^{x} + 9^{x} - 2}{2}\right)}{\dfrac{16^{x} + 9^{x} - 2}{2}}\notag\\
&= \lim_{x \to 0}\frac{16^{x} + 9^{x} - 2}{2x}\cdot 1\notag\\
&= \frac{1}{2}\lim_{x \to 0}\left(\frac{16^{x} - 1}{x} + \frac{9^{x} - 1}{x}\right)\notag\\
&= \frac{1}{2}\cdot(\log 16 + \log 9)\notag\\
&= \log\sqrt{16\cdot 9} = \log 12\notag
\end{align}
Hence $L = 12$.
