lim[(2x+1)/(2x-3)]^(1-3x)如题.求极限.底数部分应该是1吧?那么1^∞还是1?1^(-∞)还是1?

lim(x->无穷大){(2^x-3^x)/2}^(1/x) lim(x->0)((2-x)/(3-x))^1/x lim(1-x)^(2/x) x->0 1.lim x->∞ （2x/x^2+1)*cosx 2..lim x->∞{ [(x^2-x)arctanx]/x^3-x-5}3..lim x->0 （e^x-e^-x-2x)/(x-sinx) 4..lim x->0 (sinx-x)/(x^3) 5..lim x->0 (x^2*tanx)/(x-tanx) 6..lim x->0 (1+3/x)^2x 7..lim x->0 (1+2x)^(3/sinx) 8.lim x->∞ (x/1+x)^(x-3) 9..li lim(2x+3/2x+1)^(x+1) lim(2x-3/2x+1）^x+1 求极限lim(1-2/x+3/x^2)^x lim(x->2) {(3x+16)^1/2-4}/x 极限lim(x^2+2x+3)/(x-1) lim x->无限大 (1+1/2x) ^ 3x+lnx lim x-无穷 (√3x^2+1)/x+1 lim(x趋向无穷)[1-(2/x)]^3x lim x→∞(2x+1)/(3x-4) lim(x→0)2x^3-x+1 lim(1+3x)^(2/x)x趋向于零 求lim(x→0)x cos 1/x lim(x→∞)x^2/ (3x-1)的极限 lim(x-sinx)/x^3 x趋于0 不用洛必达法则～为什么不能这样做?=lim(1/x^2)-lim(sinx/x*1/x^2)=lim(1/x^2)-lim(1*1/x^2)=0 lim(x+1)^x+1(x+3)^x+3/x^2x+4 x趋于无穷