Question

$\lim_{x\to0}\frac{(729)^x-(243)^x-(81)^x+9^x+3^x-1}{x^3}$

Answer 1

6(\ln 3)^3

Answer 2

The limit expression can be rewritten using $a^x = e^{x \ln a}$. Let $k = \ln 3$. The limit becomes $\lim_{x\to0}\frac{e^{6kx}-e^{5kx}-e^{4kx}+e^{2kx}+e^{kx}-1}{x^3}$. This limit can be evaluated using the Taylor expansion of $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$ for each term. Upon expansion and collecting terms, the coefficients of $x$ and $x^2$ in the numerator are found to be zero, while the coefficient of $x^3$ is $6k^3$. Thus, the limit is $\lim_{x\to0}\frac{6k^3x^3}{x^3} = 6k^3$. Substituting $k=\ln 3$ gives the result $6(\ln 3)^3$. Alternatively, recognizing the limit as $\frac{f'''(0)}{3!}$ for $f(x) = (729)^x-(243)^x-(81)^x+9^x+3^x-1$ and calculating the third derivative at $x=0$ also yields $36(\ln 3)^3$, leading to the limit $\frac{36(\ln 3)^3}{6} = 6(\ln 3)^3$. A third method involves substituting $a=3^x$ and factoring the numerator as $(a-1)^3(a+1)(a^2+a+1)$, then using $x = \frac{\ln a}{\ln 3}$ and the standard limit $\lim_{a\to1}\frac{a-1}{\ln a}=1$ to get $6(\ln 3)^3$.