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Mathematics Question on Statistics

lim⁡x→06x−3x−2x+1x2=\displaystyle\lim_{x \to 0} \frac{6^{x} -3^{x} -2^{ x}+1}{x^{2}} =

A

log⁡e32\log_e \frac{3}{2}

B

log⁡e5\log_e 5

C

log⁡e6\log_e 6

D

log⁡e2log⁡e3\log_e 2 \log_e 3

Answer

log⁡e2log⁡e3\log_e 2 \log_e 3

Explanation

Solution

lim⁡x→06x−3x−2x+1x2\displaystyle\lim_{x \to 0} \frac{6^{x} -3^{x} -2^{ x}+1}{x^{2}}
Applying L-Hospital's rule,
=lim⁡x→06xlog⁡e6−3xlog⁡e3−2xlog⁡e22x= \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,
lim⁡x→06x(log⁡e6)2−3x(log⁡e3)2−2x(log⁡e2)22\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}
=(log⁡e6)2−(log⁡e3)2−(log⁡e2)22= \frac{ \left(\log _{e} 6\right)^{2}- \left(\log _{e} 3\right)^{2} -\left( \log _{e} 2\right)^{2}}{2}
=(log⁡e3.2)2−(log⁡e3)2−(log⁡e2)22= \frac{ \left(\log _{e} 3.2\right)^{2}- \left(\log _{e} 3\right)^{2} -\left( \log _{e} 2\right)^{2}}{2}
=(log⁡e3+log⁡e2)2−(log⁡e3)2−(log⁡e2)22=\frac{ \left(\log _{e} 3 +\log _{e} 2\right)^{2}- \left(\log _{e} 3\right)^{2} -\left( \log _{e} 2\right)^{2}}{2}
=(log⁡e3)2−(log⁡e2)2+2log⁡e3log⁡e2−(log⁡e3)2−(log⁡e2)22=\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⁡e3log⁡e2= \log _{e} 3 \log _{e} 2