Question

Evaluate: $\lim_{x \to 0} (\frac{21^x - 7^x - 3^x + 1}{x\log(1+x)})$

Answer 1

$\log 7 \log 3$

Answer 2

The limit is in $\frac{0}{0}$ form. The numerator $21^x - 7^x - 3^x + 1$ is factored as $(7^x - 1)(3^x - 1)$. The expression is then rewritten as a product of terms: $\left(\frac{7^x - 1}{x}\right) \cdot \left(\frac{3^x - 1}{x}\right) \cdot \left(\frac{x}{\log(1+x)}\right)$. Using standard limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \log a$ and $\lim_{x \to 0} \frac{\log(1+x)}{x} = 1$, the limit evaluates to $(\log 7) \cdot (\log 3) \cdot 1 = \log 7 \log 3$.