<s>[INST] 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
 [/INST] 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 $L = 12$.
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