lim((2^n+3^n)/2)^1/n（用洛必达法则怎么做啊?）

lim(n+3)(4-n)/(n-1)(3-2n) lim(n^3+n)/(n^4-3n^2+1) lim[(n+3)/(n+1))]^(n-2) 【n无穷大】 lim(2^n+3^n)^1 (n趋向无穷) lim (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中n→∞ lim根号n^2+n+1/3n-2 lim根号n^2+n+1/3n-2=? lim n->无穷大(2^n-1)/(3^n+1) 求lim n→∞ (1+2/n)^n+3 lim(n→∞)[1-(2n/n+3)] lim(3^2n+5^n)/(1+9^n)=? lim(n→∞)(2n-1/n+3) lim[(4+7+...+3n+1)/(n^2-n)]= 求1.lim(3n-(3n^2+2n)/(n-1)) 2.lim(8+1/(n+1)) 3.lim根号n(根号(n+1)-根号(n-3)) lim(1/n+2/n+3/n+4/n+5/n+……+n/n)=lim(1/n)+lim(2/n)+……+lim(n/n)成立吗?（n趋近于无穷大）为什么不成立? 高数题：n趋近于0,lim｛1/（n^2+n+1)+2/(n^2+n+2)+3/(n^2+n+3)+.+n/(n^2+n+n)｝=? lim(1/n^3+(1+2)/n^3+...+(1+2+...+n)/n^3)n趋于无穷 lim n趋于无穷大(1/n^2+3/n^2+.+2n-1/n^2