求(1+2^x+3^x)^1/x当x趋于正无穷时的极限?
1个回答
展开全部
3^x < 1+2^x+3^x < 3 * 3^x
3 < (1+2^x+3^x) ^ (1/x) < 3^(1/x) * 3
lim(x->+∞) 3^(1/x) = 1
由迫敛准则(夹挤准则)得:
lim(x->+∞) (1+2^x+3^x) ^ (1/x) = 3,9,∵lim(x->+∞)[ln(1+2^x+3^x)/x]
=lim(x->+∞)[(ln(1+2^x+3^x))'/(x)'] (∞/∞型极限,应用罗比达法则)
=lim(x->+∞)[(2^x*ln2+3^x*ln3)/(1+2^x+3^x)]
=lim(x->+∞)[((2/3)^x*ln2+ln3)/...,3,先取对数,再用洛比达法则,lim [ln(1+2^x+3^x)]/x=lim(2^xln2+3^xln3)/(1+2^x+3^x)=ln3
所以lim(1+2^x+3^x)^1/x=e^(ln3)=3 (x趋于正无穷),1,
3 < (1+2^x+3^x) ^ (1/x) < 3^(1/x) * 3
lim(x->+∞) 3^(1/x) = 1
由迫敛准则(夹挤准则)得:
lim(x->+∞) (1+2^x+3^x) ^ (1/x) = 3,9,∵lim(x->+∞)[ln(1+2^x+3^x)/x]
=lim(x->+∞)[(ln(1+2^x+3^x))'/(x)'] (∞/∞型极限,应用罗比达法则)
=lim(x->+∞)[(2^x*ln2+3^x*ln3)/(1+2^x+3^x)]
=lim(x->+∞)[((2/3)^x*ln2+ln3)/...,3,先取对数,再用洛比达法则,lim [ln(1+2^x+3^x)]/x=lim(2^xln2+3^xln3)/(1+2^x+3^x)=ln3
所以lim(1+2^x+3^x)^1/x=e^(ln3)=3 (x趋于正无穷),1,
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询