lim(x→0)(e∧x+x)∧1/x

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lim(x→0)(e∧x+x)∧1/xlim(x→0)(e∧x+x)∧1/xlim(x→0)(e∧x+x)∧1/xx→0时lim(e^x+x)^(1/x)=lime^ln(e^x+x)^(1/x)=e

lim(x→0)(e∧x+x)∧1/x
lim(x→0)(e∧x+x)∧1/x

lim(x→0)(e∧x+x)∧1/x
x→0时lim(e^x+x)^(1/x)=lime^ln(e^x+x)^(1/x)=e^limln(e^x+x)^(1/x),而limln(e^x+x)^(1/x)=lim[ln(e^x+x)]/x,
用洛必塔法则lim [ln(e^x+x)] / x= lim [ln(e^x+x)]' / x'=lim (e^x+1) / (e^x+x)=(1+1) / (1+0)=2,
故原式=e^2