Question

limx趋于无穷，{（π/2）arctanx}的（1/x）次方，用凑e法求极限，怎么求，步骤

Answer 1

问：limx趋于无穷，{（π/2）arctanx}的（1/x）次方，用凑e法求极限，怎么求，步骤

答：您好 很高兴为您解答，根据您的提问？我的解答是：步骤一：用凑e法求极限首先，将θ=π/2式子化简。（π/2）Arctan x = (π/2) * (ln (1+ x^2))/2然后，令 y=x^2(π/2)Arctan x = (π/2) * (ln (1+ y))/2最后，让y→∞ ,即求取以下极限lim y→∞ {(π/2)* (ln (1+ y))/2}/y^(1/x)步骤二：用凑e法求极限可以看到，y的指数比该分子项低一个次数，我们先用凑e法将该分子项表达为e的指数：lim y→∞ {(π/2)* (ln (1+ y))/2}/y^(1/x) = lim y→∞ (π/2) * (e^(ln (1+ y))/2)/y^(1/x)=lim y→∞ (π/2) * (1+y)^(1/2)/y^(1/x)由于该分数项可以看出，y的指数会失去，故最后极限值为：lim y→∞ (π/2) * (1+y)^(1/2)/y^(1/x) = lim y→∞ (π/2) * ∞ = +∞

问：题都不对着

答：您好根据您的提问limx趋于无穷，{（π/2）arctanx}的（1/x）次方，用凑e法求极限，怎么求，步骤？我的解答是：lim(x->正无穷)(π/2-arctanx)/(1/x)用洛必达法则 上下求导得lim(x->正无穷)(-1/(1+x^2))/(-1/x^2)=lim(x->正无穷)x^2/(1+x^2)=lim(x->正无穷)1/(1+1/x^2)=1