Question

一道高数极限题求助

求解这两种情况的极限如何求

Answer 1

简单计算一下即可，答首肢芹历案如者首世图所示

Answer 2

a>1

lim(x->+无穷) [x+√(x^2+4)]/(x+2)^a

分子最高阶=1

分母最高阶=a>1

=>

lim(x->+无穷碧掘) [x+√(x^2+4)]/(x+2)^a =0

a<1

lim(x->+无穷) [x+√(x^2+4)]/(x+2)^a

分子最高阶=1

分母最高阶唯扰=a<1

=>

lim(x->+无穷) [x+√(x^2+4)]/(x+2)^a  ->+无指慧旦穷

Answer 3

a > 1 时， lim<x→+∞>[x+√神滚核(x^2+4)]/游掘(x+2)^a
= lim<x→+∞>[x+√(x^2+4)]/[(x+2)(x+2)^(a-1)]
= lim<x→+∞>{[1+√(1+4/x^2)]/(1+2/x)}[1/(x+2)^(a-1)] = 0
a < 1 时， lim<x→+∞>[x+√(x^2+4)]/(x+2)^a
= lim<x→+∞>[x+√(x^2+4)](x+2)^(1-a)/备如(x+2)
= lim<x→+∞>[1+√(1+4/x^2)](x+2)^(1-a)/(1+2/x) = +∞