Question

$\underset{x\to 0}{lim}\frac{{27}^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=$

• A. $0$
• B. $8\sqrt{2}(\log 3{)}^2$
• C. $8(\log 3{)}^2$
• D. $1$

Answer 1

The correct option is B $8\sqrt{2}(\log 3{)}^2$

$\underset{x\to 0}{lim}\frac{{27}^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$
$={lim}_{x\to 0}\frac{({27}^x-9^x-3^x+1)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$
$={lim}_{x\to 0}\frac{\left[9^x\right(3^x-1)-1(3^x-1\left)\right](\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$
$={lim}_{x\to 0}\frac{(9^x-1)(3^x-1)(\sqrt{2}+\sqrt{1+\cos x})}{2{\sin }^2(\frac{x}{2})}$
$={lim}_{x\to 0}\frac{\frac{(9^x-1)}{x}\frac{(3^x-1)}{x}(\sqrt{2}+\sqrt{1+\cos x})}{\frac{2{\sin }^2(\frac{x}{2})}{{4\left(\frac{x}{2}\right)}^2}}$
$=\frac{\log 9\log 3\left(2\sqrt{2}\right)}{1/2}=8\sqrt{2}{\left(\log 3\right)}^2$