高数洛必达法则的一道题
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lim(x→∞) x[(1+1/x)^x-e]
u=1/x
=lim(u→0)[(1+u)^(1/u)-e]/u
lim(u→0)(1+u)^(1/u)=e
=lim(u→0)[(1+u)^(1/u)-e]'/u'
y= (1+u)^(1/u)
lny=(1/u)ln(1+u)
(lny)'=(-1/u^2)ln(1+u)+(1/u)/(1+u)
y'/y=ln(1+u)/-u^2 +1/(u^2+u))
y'=(1+u)^(1/u)*[ln(1+u)/-u^2+1/(u^2+u)]
[(1+u)^(1/u)]'=(1+u)^(1/u) *[ln(1+u)/-u^2 +1/(u^2+u)]
lim(x→∞)x[(1+1/x)^x-e]
=lim(u→0) (1+u)^(1/u)* [ln(1+u)/-u^2 +1/(u^2+u)]
lim(u→0)(1+u)^(1/u)=e
lim(u→0) ln(1+u)/-u^2=0
lim(u→0)1/(u^2+u)→∞
=∞
u=1/x
=lim(u→0)[(1+u)^(1/u)-e]/u
lim(u→0)(1+u)^(1/u)=e
=lim(u→0)[(1+u)^(1/u)-e]'/u'
y= (1+u)^(1/u)
lny=(1/u)ln(1+u)
(lny)'=(-1/u^2)ln(1+u)+(1/u)/(1+u)
y'/y=ln(1+u)/-u^2 +1/(u^2+u))
y'=(1+u)^(1/u)*[ln(1+u)/-u^2+1/(u^2+u)]
[(1+u)^(1/u)]'=(1+u)^(1/u) *[ln(1+u)/-u^2 +1/(u^2+u)]
lim(x→∞)x[(1+1/x)^x-e]
=lim(u→0) (1+u)^(1/u)* [ln(1+u)/-u^2 +1/(u^2+u)]
lim(u→0)(1+u)^(1/u)=e
lim(u→0) ln(1+u)/-u^2=0
lim(u→0)1/(u^2+u)→∞
=∞
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