Question

${lim}_{x\to 0}\frac{{\left(729\right)}^x-{\left(243\right)}^x-{\left(81\right)}^x+9^x+3^x-1}{x^3}$

Answer 1

Given:

$\underset{x\to 0}{lim}\frac{(729{)}^x-(243{)}^x-(81{)}^x+9^x+3^x-1}{x^3}$

$\underset{x\to 0}{lim}\frac{3^{6x}-3^{5x}-3^{4x}+3^{2x}+3^x-1}{x^3}$

$\underset{x\to 0}{lim}\frac{3^{5x}(3^x-1)-3^{2x}\left[\right(3^x-1\left)\right(3^x+1)+1(3^{}x)-1]}{x^3}$

$\underset{x\to 0}{lim}\frac{\left(\frac{3^x-1}{x}\right)[3^{5x}-3^{3x}-3^{2x}+1]}{x^2}$

$\underset{x\to 0}{lim}\left(\frac{3^x-1}{n}\right)\times \underset{x\to 0}{lim}\left[\frac{3^{3x}(3^{2x}-1)-1(3^{2x}-1)}{x^2}\right]$

${\ln }_3\times \underset{x\to 0}{lim}\frac{\left[\right(3^{3x}-1\left)\right(3^{2x}-1\left)\right]}{x^2}$

${\ln }_3\times \underset{x\to 0}{lim}\frac{3^x-1}{x}(3^x+1)\frac{(3^x-1)}{x}(3^{2x}+1+3^4)$

${\ln }_3\times \left(\ln 3\right)(3^0+1)\left({\ln }_3\right)(3^0+1+3^o)$

$\left(\ln 3{)}^3\right(2\left)\right(3)$

$=6(\ln 3{)}^3\{here{\ln }_3={\log }_{e}3\}$