lim(n→∞)[5^n-4^(n-1)]/[5^(n+1)+3^(n+2)]可是计算过程太简略了，看不懂哎

用数列极限证明lim(n→∞)(n^-2)/(n^+n+1)=1中证明如下:lim(n→∞)3n+1/5n-4 lim {（-4）^n+5^n}/{4^(n+1) + 5^(n+1)} n→无限 lim n→∞ [(n+1)^4/5^(n+1)]/[n^4/5^n]为何等于1/5, lim((5^n-4^(n-1))/((5^(n+1)+3^(n+2)) n→∞时的极限是多少? 求lim(n→∞)时[5^n-4^(n-1)]/5^(n+1)+3^(n+2) 求极限 lim(n→∞) tan^n (π/4 + 2/n) lim(n→∞)tan^n(π/4+2/n) =lim(n→∞)[(tan(π/4)+tan(2/n))/(1-tan(π/4)tan(2/n))]^n =lim(n→∞)[(1+tan(2/n))/(1-tan(2/n))]^n =lim(n→∞)(1+tan(2/n))^n/(1-tan(2/n))^n (1) 因为 lim(n→∞)(1+tan(2/n) lim(n→∞) 1/(n+1)-2/(n+1)+3/(n+1)-4/(n+1)+...+[(2n-1)/(n+1)]-[(2n)/(n-1)]求极限 大一微积分解答：lim(n→+∞)（2^n+4^n+6^n+8^n）^1/n=? lim (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中n→∞ lim（n→∞) （(2n!/n!*n)^1/n的极限用定积分求是lim（n→∞) 1/n(2n!/n!)^1/n 不好意思 lim(n→∞)3n^2+5n-7/4-n^2的值是 求当n→∞,Lim(1+2+3+4+……+(n-1)+n)/n lim[n→∞] (x^n+1)^(1/n) 求lim n→∞ (1+2/n)^n+3 lim(n→∞)[1-(2n/n+3)] lim(n→∞)(2n-1/n+3) 求 lim (n→+∞) n^( 1/n)的极限 lim(arctan n)^1/n (n→∞)求极值