Question

$\text{If}\underset{x\to 0}{lim}\frac{{64}^x-{32}^x-{16}^x+4^x+2^x-1}{(\sqrt{8+\cos x}-3)\sin x}=a(\ln 2{)}^b,$
then the values of $a$ and $b$ are

• A. $a=-36,b=3$
• B. $a=-72,b=2$
• C. $a=-72,b=3$
• D. $a=-36,b=2$

Answer 1

The correct option is C $a=-72,b=3$

$L=\underset{x\to 0}{lim}\frac{(2^x-1)(4^x-1)(8^x-1)}{\left(\frac{8+\cos x-9}{\sqrt{8+\cos x}+3}\right)\sin x}=\underset{x\to 0}{lim}\frac{(2^x-1)(4^x-1)(8^x-1)}{\frac{-(1-\cos x)\left(\sin x\right)(1+\cos x)}{1+\cos x}}\times (\sqrt{8+\cos x}+3)=\underset{x\to 0}{lim}\frac{(2^x-1)(4^x-1)(8^x-1)}{-{\sin }^{3}x}\times (1+\cos x)(\sqrt{8+\cos x}+3)=\underset{x\to 0}{lim}\frac{\left(\frac{2^x-1}{x}\right)\left(\frac{4^x-1}{x}\right)\left(\frac{8^x-1}{x}\right)}{-(\frac{\sin x}{x}{)}^3}\times (1+\cos x)(\sqrt{8+\cos x}+3)=\frac{\ln 2.\ln 4.\ln 8}{-1}\times (1+1)(\sqrt{8+1}+3)=-\left(\ln 2\right)2\left(\ln 2\right)3\left(\ln 2\right)\times 2\times 6=-72(\ln 2{)}^3\Rightarrow a=-72,b=3$