Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{(72{)}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}:& x\ne 0\\ k\log 2\log 3;& x=0\end{array}$.
If $f\left(x\right)$ is continuous function at $x=0$, then $\sqrt{2}k=$

Answer 1

Given function is
$f\left(x\right)=\{\begin{array}{cc}\frac{(72{)}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}:& x\ne 0\\ k\log 2\log 3;& x=0\end{array}$
$f\left(x\right)$ is continuous at $x=0$
$\Rightarrow \underset{x\to 0}{\text{lim}}f\left(x\right)=f\left(0\right)$
$f\left(0\right)=\underset{x\to 0}{lim}\frac{(72{)}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$
$=\underset{x\to 0}{lim}\frac{(9^x-1)(8^x-1)}{\sqrt{2}-\sqrt{2}\cos \frac{x}{2}}[\text{as}x\to 0]$
$=\underset{x\to 0}{lim}\frac{9^x-1}{x}\times \frac{8^x-1}{x}\times \frac{x^2}{\sqrt{2}-\sqrt{2}\cos \frac{x}{2}}$
$=\log 9\cdot \log 8\cdot \underset{x\to 0}{lim}\frac{2x}{\frac{\sqrt{2}}{2}\sin \frac{x}{2}}$
$=6\log 3\cdot \log 2\cdot 4\sqrt{2}$
$=24\sqrt{2}\log 2\cdot \log 3$
$\therefore k=24\sqrt{2}$