Question

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

• A. $8\sqrt{2}(\log 3{)}^2$
• B. $8\sqrt{2}(\log 3{)}^3$
• C. $4\sqrt{2}(\log 3{)}^3$
• D. $4\sqrt{2}(\log 3{)}^2$

Answer 1

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

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

$=2\sqrt{2}.\underset{x\to 0}{lim}\frac{(9^x-1)(3^x-1)}{1-\cos x}=2\sqrt{2}.\underset{x\to 0}{lim}\frac{\left(\frac{9^x-1}{x}\right)\cdot \left(\frac{3^x-1}{x}\right)}{\frac{1-\cos x}{x^2}}$

$=2\sqrt{2}\log 9\cdot \log 3\cdot 2=8\sqrt{2}(\log 3{)}^2$