Lim =(2x-3)/（x^2-5x=4）= 极限等于多少啊!x--1

lim(x→∞)[(X^2+2x+5)/(3x^2-x+1) = 数学难题,求指教~~极限到底是什么玩意我还不懂.1.lim（x^2-2x+6）2.lim(x^2-4)/(x-2）3.lim(x^2-6x=8)/(x^-5x=4)4.lim[1/(1-x)-1/(1-x^3)5.lim(sin2x)/x^26.lim(sin5x)/x7.lim(tan3x)/(sin7x)8.lim[(1+x)/x]^2x例如这道例题~lim[(1/3)x+1] Lim(x→∞)((2x-3k)/2x)-2x= Lim(x→∞)xsin3/x, lim[x→1]f(x)存在,且f(x)=2x+5+3lim[x→1]f(x),求f(x) 洛必达法则(1)条件?(2)lim(x->0)P(x)/Q(x)=?已知:lim(x->0)P(x)=A(x),lim(x->0)Q(x)=B(x)可不可以:lim(x->0)P(x)/Q(x)=lim(x->0)A(x)/B(x)(3)lim(x->0)[(1+x)^(1/x)-e]/(cosx-1)=? 为什么lim(x^3-2x+5)/(3x^5+2x+3)=0 lim(x^3+2x+a)/(x-1)=5 求a ( x趋近于1) lim(x^2+1/x^2-2x)^x=? lim(x/x-c)^x=2 x趋于无穷 lim(x+1/x+2)^x=?x趋于无穷 lim(x趋于无穷)(x/1+x)^(x+2)=? lim(x->无穷大)x/(sinx)^2=? 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(3*x^2+5)/(5x+3)sin(2/x)=多少?lim(3*x^2+5)/(5x+3)*sin(2/x)=多少？ lim[f(2x)/x]=1/3 则 lim[x/f(3x)]= (x-0) lim x趋于0 f(x)/x^2=5 求lim x趋于0 f(x)=? 已知lim(x→0）[f(3x)/x]=3 求lim(x→0) [2x/f(5x)] 设lim（x→0）[f(x)-3]/x^2=100,求lim(x→0)f(x)