f ( x ) = 72 x − 9 x − 8 x + 1 2 − 1 + cos ⁡ x , x ≠ 0 k log ⁡ 2 log…

Given that $f\left(x\right)=\left\{\begin{array}{cl}\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}},& x\ne 0\\ k\log 2\log 3,& x=0\end{array}\right.$

We know that $f\left(x\right)$ is continuous $\Rightarrow \underset{x\to a}{\lim }f\left(x\right) = f\left(a\right)$

$\underset{x\to 0}{\lim }\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} =k \log \left(2\right)\log \left(3\right)$

$\underset{x\to 0}{\lim }\frac{\left(9^x-1\right)\left(8^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}\times \frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}} =k \log \left(2\right)\log \left(3\right)$

$\underset{x\to 0}{\lim }\frac{\left(9^x-1\right)\left(8^x-1\right)}{2-1-\cos x}\times \frac{\sqrt{2}+\sqrt{1+\cos x}}{1} =k \log \left(2\right)\log \left(3\right)$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(9^x-1\right)\left(8^x-1\right)}{2{\sin }^2\left({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}\times \frac{\sqrt{2}+\sqrt{1+\cos x}}{1} =k \log \left(2\right)\log \left(3\right)$

We know that $\underset{x\to 0}{\lim }\frac{\sin ax}{x} = a , \underset{x\to 0}{\lim }\frac{a^x-1}{x}= \log \left(a\right)$

$\Rightarrow \underset{x\to 0}{\lim } \frac{x^2}{2{\sin }^2\left({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}\times \underset{x\to 0}{\lim } \frac{\left[9^x-1\right]}{x}\times \underset{x\to 0}{\lim } \frac{\left[8^x-1\right]}{x}\underset{ x\to 0}{\times \lim } \left[\sqrt{2}+\sqrt{1+\cos x}\right]=k \log \left(2\right)\log \left(3\right)$

$\Rightarrow \frac{1}{2\times {\displaystyle \frac{1}{4}}}\times \log \left(3^2\right)\times \log \left(2^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\times 2\sqrt{2} = k \log \left(3\right)\times \log \left(2\right)$

$\Rightarrow 2\times 2\log \left(3\right)\times \frac{3}{2}\log \left(2\right)\times 2\sqrt{2} = k \log \left(3\right)\times \log \left(2\right)$

$\left(24\sqrt{2}\right)\left[\log \left(3\right)\times \log \left(2\right)\right]= k \log \left(3\right)\times \log \left(2\right)$

So required value $k =24\sqrt{2.}$