Question

If $\underset{x\to 0}{lim}\frac{{64}^x-{32}^x-{16}^x+4^x+2^x-1}{\left[\sqrt{(15+\cos x)-4}\right]\sin x}=-a{\left(\log b\right)}^3.$ Find $a+b$.

Answer 1

If $2^x=t$, then $N^r=t^6-t^5-t^4+t^2+t-1$
$=t^5(t-1)-t^2(t^2-1)+(t-1)$
$=(t-1)[t^5-t^2(t+1)+1]$
$=(t-1)(t^5-t^3-t^2+1)$
$=(t-1)(t^3-1)(t^2-1)$
$=(2^x-1)(2^{3x}-1)(2^{2x}-1)$
Rationalizing, we have to evaluate $\stackrel{Lt}{x\to 0}\frac{2^x-1}{x}.\frac{2^{2x}-1}{2x}.\frac{2^{3x}-1}{3x}.6.\frac{x^3(\sqrt{15+\cos x}+4)}{(\cos x-1)\sin x}$
$=\log 2.\log 2.\log \mathrm{2.6.}\frac{x^3(\sqrt{15+\cos x}+4)}{-2{\sin }^2\left(x/2\right).\sin x}$
$=-6{\left(\log 2\right)}^3.\frac{x^2}{2{\sin }^2\left(x/2\right)}.\frac{x}{\sin x}.8$
$=-6{\left(\log 2\right)}^3.\frac{x^2}{2{\left(x/2\right)}^2}.8=-96{\left(\log 2\right)}^3.$
Ans: 98