Question

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

• A. $0$
• B. $8\sqrt{2}(\log 3{)}^2$
• C. $8\left(\log 3{)}^{2}4\right)1$
• D. None of these

Answer 1

Answer: (A)

$0$

$⟹{lim}_{x\to 0}\frac{{27}^x-9^x-3^x+1}{\sqrt{2}(1-\cos \frac{x}{2})}$

$⟹{lim}_{x\to 0}\frac{{27}^x-9^x-3^x+1}{2\sqrt{2}{\sin }^2\frac{x}{4}}$

Using L hospitals rule

${lim}_{x\to 0}\frac{{27}^x\log 27-9^x\log 9-3^x\log 3}{2\sqrt{2}\sin \frac{x}{2}}$

${lim}_{x\to 0}\frac{{27}^x(\log 27{)}^2-9^x(\log 9{)}^2-3^x(\log 3{)}^2}{\sqrt{2}\cos \frac{x}{2}}$

By substituting limits

$\frac{9(\log 3{)}^2-4(\log 3{)}^2-(\log 3{)}^2}{\sqrt{2}}=2\sqrt{2}(\log 3{)}^2\approx 0.64$