帮忙求几个未定式的极限.1 .lim(x→0)[(a^1/x + b^1/x)/2]^2x2.lim(x→0)(1/sin^2X - 1/x^2)1.ab 2.1/3

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帮忙求几个未定式的极限.1.lim(x→0)[(a^1/x+b^1/x)/2]^2x2.lim(x→0)(1/sin^2X-1/x^2)1.ab2.1/3帮忙求几个未定式的极限.1.lim(x→0)[

帮忙求几个未定式的极限.1 .lim(x→0)[(a^1/x + b^1/x)/2]^2x2.lim(x→0)(1/sin^2X - 1/x^2)1.ab 2.1/3
帮忙求几个未定式的极限.
1 .lim(x→0)[(a^1/x + b^1/x)/2]^2x
2.lim(x→0)(1/sin^2X - 1/x^2)
1.ab 2.1/3

帮忙求几个未定式的极限.1 .lim(x→0)[(a^1/x + b^1/x)/2]^2x2.lim(x→0)(1/sin^2X - 1/x^2)1.ab 2.1/3
2.1/(sinx)^2 - 1/x^2 = [x^2 - (sinx)^2]/[xsinx]^2
= [x^2 - (sinx)^2]/x^4*x^2/[sinx]^2
lim_{x->0}[1/(sinx)^2 - 1/x^2]
= lim_{x->0}[x^2 - (sinx)^2]/x^4*x^2/[sinx]^2
= lim_{x->0}[x^2 - (sinx)^2]/x^4
= lim_{x->0}[2x - 2sin(x)cos(x)]/[4x^3]
= lim_{x->0}[2x - sin(2x)]/[4x^3]
= lim_{x->0}[2 - 2cos(2x)]/[12x^2]
= lim_{x->0}[1 - cos(2x)]/[6x^2]
= lim_{x->0}[2sin(2x)]/[12x]
= 2/6lim_{x->0}[sin(2x)]/[2x]
= 2/6
=1/3