Question

The value of $\underset{x\to 0}{lim}\frac{{27}^x-9^x-3^x+1}{\sqrt{5}+\sqrt{4+\cos x}}$ is

• A. $\sqrt{5}{\left(\log 3\right)}^2$
• B. $16\sqrt{5}{\left(\log 3\right)}^{}$
• C. $8\sqrt{5}{\left(\log 3\right)}^{}$
• D. $8\sqrt{5}{\left(\log 3\right)}^2$

Answer 1

The correct option is D $8\sqrt{5}{\left(\log 3\right)}^2$

Let $L=\underset{x\to 0}{lim}\frac{{27}^x-9^x-3^{x+1}}{\sqrt{5}-\sqrt{4+cosx}}$

$=\underset{x\to 0}{lim}\frac{9^x.3^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+cosx}}$

$=\underset{x\to 0}{lim}\frac{9^x(3^x-1)-1(3^x-1)}{\sqrt{5}-\sqrt{4+cosx}}$

$=\underset{x\to 0}{lim}\frac{9^x(3^x-1)-1(3^x-1)}{\sqrt{5}-\sqrt{4+cosx}}\times \frac{\sqrt{5}+\sqrt{4+cosx}}{\sqrt{5}+\sqrt{4+cosx}}\times \frac{x^2}{x^2}$

$=\underset{x\to 0}{lim}\frac{(9^x-1)(3^x-1)}{x.x}\times \frac{x^2}{(5-4-cosx)}(\sqrt{5}+\sqrt{4+cosx})$

$=ln9.ln3\times 2\sqrt{5}\times \underset{x\to 0}{lim}\frac{x^2}{1-cosx}\dots (\underset{x\to 0}{lim}=\frac{a^x-1}{x}=\ln a)$

$=ln9.ln3\times 2\sqrt{5}\times \underset{x\to 0}{lim}\frac{x}{2si{n}^{2}x/2}$

$=ln{3}^2.ln3\times 2\sqrt{5}\times 2\times \underset{x\to 0}{lim}{\left[\frac{\frac{x}{2}}{sinx/2}\right]}^2$

$=4\sqrt{5}\times 2\times ln3\times ln3$

$L=8\sqrt{5}(log3{)}^2$