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利用等价无穷小
(e^x)-1 ~ x,ln(1+x) ~ x,1-cosx ~ (x^2)/2 (x→0),
可得
lim(x→0){[(2+cosx)/3]^x-1}/x^3
= lim(x→0){e^{xln[(2+cosx)/3]}-1}/x^3
= lim(x→0){xln[(2+cosx)/3]}/x^3
= lim(x→0){ln[(2+cosx)/3]}/x^2
= lim(x→0){ln[1+(cosx-1)/3]}/x^2
= lim(x→0)[(cosx-1)/3]/x^2
= (-1/3)lim(x→0)[(1-cosx)/(x^2)]
= (-1/3)*(1/2)
= -1/6。
(e^x)-1 ~ x,ln(1+x) ~ x,1-cosx ~ (x^2)/2 (x→0),
可得
lim(x→0){[(2+cosx)/3]^x-1}/x^3
= lim(x→0){e^{xln[(2+cosx)/3]}-1}/x^3
= lim(x→0){xln[(2+cosx)/3]}/x^3
= lim(x→0){ln[(2+cosx)/3]}/x^2
= lim(x→0){ln[1+(cosx-1)/3]}/x^2
= lim(x→0)[(cosx-1)/3]/x^2
= (-1/3)lim(x→0)[(1-cosx)/(x^2)]
= (-1/3)*(1/2)
= -1/6。
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