calculate limit

Question

$\lim_{x\rightarrow 0} \frac{6^{x}-2^{x}-3^{x}+1}{x^{2}}$

Answer 1

$\lim_{ x\to 0} \frac{(6^x-2^x-3^x+1)}{x^2}  \\= \lim_{x \to 0}  \frac{ [(2.3)^x-2^x-3^x+1] }{x^2} \\ = \lim_{x \to 0} \frac{ [(3^x-1)(2^x-1)]} {x^2} \\= \lim_{x \to 0} \frac{3^x-1}{x}  \times \lim_{ x \to 0} \frac{2^x-1}{x} \\= \ln 3 . \ln 2$.

Answer 2

lt x->0   ( 6^x-2^x-3^x+1)/x^2

=(6^xln6-2^xln2-3^xln3)/2x   using l-hospital

=(6^x(ln6)^2-2^x(ln2)^2-3^x(ln3)^2)/2   again using l-hospital

=((ln6)^2-(ln2)^2-(ln3)^2)/2

=((ln(2*3))^2-(ln2)^2-(ln3)^2)/2

=(2*ln2*ln3/2)=ln2*ln3