求极限 lim(n→∞) tan^n (π/4 + 2/n)

题目

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))^n
=lim(n→∞){[1+tan(2/n)]^(1/(tan(2/n))}^[2(tan(2/n)/(2/n)]
=e^2,(2)
lim(n→∞)(1-tan(2/n))^n
=lim(n→∞){[1-tan(2/n)]^(-1/(tan(2/n))}^[-2(tan(2/n)/(2/n)]
=e^(-2),(3)
由(1),(2),(3)得
lim(n→∞)tan^n(π/4+2/n)=e^2/e^(-2)=e^4.
其中第一步就看不懂了：
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

答案

解 lim(n→∞)tan^n(π/4+2/n)
=lim(n→∞)[(tan(π/4)+tan(2/n))/(1-tan(π/4)tan(2/n))]^n（着一步就是tan（A+B）拆分的公式啊~
=lim(n→∞)[(1+tan(2/n))/(1-tan(2/n))]^n（着一步就是求出tan（π/4）=1啊
哪里看不明白再追问