第1个回答 2018-12-14
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4610b912c8fcc3ce7a8ef88c9f45d688d53f20c6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
可以不用洛必达,其中用到有理化
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第2个回答 2011-10-06
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
第3个回答 2011-10-06
解:∵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)/((1/3)^x+(2/3)^x+1)]
=(0*ln2+ln3)/(0+0+1)
=ln3
∴lim(x->+∞)[(1+2^x+3^x)^(1/x)]
=lim(x->+∞){(e^[ln(1+2^x+3^x)/x]}
=e^{lim(x->+∞)[ln(1+2^x+3^x)/x]}
=e^3。
第4个回答 2011-10-06
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)