Question

当x趋近于正无穷时，求lim[x+根号（1+x^2)]^1/x的极限

Answer 1

解：∵lim(x->+∞)[ln(x+√(1+x^2))/x]
=lim(x->+∞)[1/√(1+x^2)]
(∞/∞型极限，应用罗比达法则)
=0
∴lim(x->+∞)[(x+√(1+x^2))^(1/x)]
=lim(x->+∞){e^[ln(x+√(1+x^2))/x]}
=e^{lim(x->+∞)[ln(x+√(1+x^2))/x]}
=e^0
=1。

Answer 2

求当x趋近于正无穷大时lim[x+1/(x-2)]^x的极限值?
解：x→+∞lim[x+1/(x-2)]^x=x→+∞lim[(x²-2x+1)/(x-2)]^x=x→+∞lim[(x-
2+1/x)/(1-2/x)]^x=+∞
其中分母(1-2/x)→1，分子(x-2+1/x)→+∞.
如果分子是(x+1),则：
x→+∞lim[(x+1)/(x-2)]^x=x→+∞lim[1+3/(x-2)]^x
=x→+∞lim{[1+3/(x-2)]^[(x-2)/3]}³{[1+3/(x-2)]²}
=x→+∞lim{[1+3/(x-2)]^[(x-2)/3]}³{x→+∞lim[1+3/(x-2)]²}=e³