Question

$\displaystyle\lim_{x \to 0} \frac{6^{x} -3^{x} -2^{ x}+1}{x^{2}} = $

• A. $\log_e \frac{3}{2}$
• B. $\log_e 5$
• C. $\log_e 6$
• D. $\log_e 2 \log_e 3$

Answer 1

Answer: (D)

$\displaystyle\lim_{x \to 0} \frac{6^{x} -3^{x} -2^{ x}+1}{x^{2}} $
Applying L-Hospital's rule,
$= \displaystyle\lim _{x\to 0} \frac{6^{x} \log_{e} 6-3^{x } \log _{e} 3-2^{x} \log _{e} 2}{2x} $
Again, applying L-Hospital's rule,
$ \displaystyle\lim _{x\to 0} \frac{6^{x} \left(\log _{e} 6\right)^{2}-3^{x } \left(\log _{e} 3\right)^{2} - 2^{x}\left( \log _{e} 2\right)^{2}}{2}$
$= \frac{ \left(\log _{e} 6\right)^{2}- \left(\log _{e} 3\right)^{2} -\left( \log _{e} 2\right)^{2}}{2} $
$= \frac{ \left(\log _{e} 3.2\right)^{2}- \left(\log _{e} 3\right)^{2} -\left( \log _{e} 2\right)^{2}}{2} $
$=\frac{ \left(\log _{e} 3 +\log _{e} 2\right)^{2}- \left(\log _{e} 3\right)^{2} -\left( \log _{e} 2\right)^{2}}{2} $
$=\frac{ \left(\log _{e} 3\right)^{2}- \left(\log _{e} 2\right)^{2} + 2 \log _{e} 3 \log _{e} 2 - \left(\log _{e} 3\right)^{2} - \left( \log _{e} 2\right)^{2}}{2} $
$= \log _{e} 3 \log _{e} 2$