0 votes 0 votes $\lim_{x\rightarrow 0} \frac{6^{x}-2^{x}-3^{x}+1}{x^{2}}$ Calculus calculus limits + – saket nandan asked Jul 10, 2015 • retagged Aug 2, 2015 by Arjun saket nandan 336 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes $\lim_{ x\to 0} \frac{(6^x-2^x-3^x+1)}{x^2} \\= \lim_{x \to 0} \frac{ [(2.3)^x-2^x-3^x+1] }{x^2} \\ = \lim_{x \to 0} \frac{ [(3^x-1)(2^x-1)]} {x^2} \\= \lim_{x \to 0} \frac{3^x-1}{x} \times \lim_{ x \to 0} \frac{2^x-1}{x} \\= \ln 3 . \ln 2$. saket nandan answered Jul 12, 2015 • selected Jul 12, 2015 by Arjun saket nandan comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes 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 Sudeep Choudhary answered Jul 20, 2015 Sudeep Choudhary comment Share Follow See all 0 reply Please log in or register to add a comment.