Question

$\displaystyle \lim _{x \rightarrow 0} \frac{6^{x}-3^{x}-2^{x}+1}{x^{2}}$ is equal to

• A. $\left(\log _{e} 2\right) \log _{e} 3$
• B. $\log _{e} 5$
• C. $\log _{e} 6$
• D. 0

Answer 1

Answer: (A)

We have,
$\displaystyle\lim _{x \rightarrow 0} \frac{6^{x}-3^{x}-2^{x}+1}{x^{2}}=\lim _{x \rightarrow 0} \frac{3^{x} \cdot 2^{x}-3^{x}-2^{x}+1}{x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{3^{x}\left(2^{x}-1\right)-1\left(2^{x}-1\right)}{x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{3^{x}-1}{x} \times \displaystyle\lim _{x \rightarrow 0} \frac{2^{x}-1}{x}$
$=\left(\log _{e} 2\right)\left(\log _{e} 3\right)$