Question

高数，如图？

Answer 1

如图所示，一直用洛必达法则就可以了

Answer 2

lim(n->∞) ln(e^n+x^n+sinn)/n
consider
lim(y->∞) ln(e^y+x^y+siny)/y (∞/∞ 分子分母分别求导)
=lim(y->∞) (e^y + lnx. x^y + cosy) /(e^y+x^y+siny)
case 1: 0<x< e
lim(y->∞) (e^y + lnx. x^y + cosy) /(e^y+x^y+siny)
分子分母同时除以 e^y
=lim(y->∞) [ 1 + lnx .(x/e)^y + cosy/e^y ] /[1+(x/e)^y+siny/e^y'
=( 1+0+0) /(1+0+0)
=1
case 2 : x=e
lim(y->∞) (e^y + lnx. x^y + cosy) /(e^y+x^y+siny)
=lim(y->∞) (e^y + e^y + cosy) /(e^y+e^y+siny)
=lim(y->∞) (2e^y + cosy) /(2e^y+siny)
分子分母同时除以 e^y
=lim(y->∞) (2 + cosy/e^y) /(2+siny/e^y)
=(2+0)/(2+0)
=1
case 3: x>e
lim(y->∞) (e^y + lnx. x^y + cosy) /(e^y+x^y+siny)
分子分母同时除以 x^y
=lim(y->∞) [(e/x)^y + lnx.+ cosy/x^y ] /[(e/x)^y+1+siny/x^y ]
=(0+lnx+0)/(0+1+0)
=lnx
ie
lim(n->∞) ln(e^n+x^n+sinn)/n
=1 ; 0<x≤ e
=lnx ; x>e