这道求极限的题怎么做呀?思路是什么呢?
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将三角函数展成泰勒级数,根据分母可知,只需取级数的前两项,剩余项均为可去余项,接下来就很简单了。
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这个应该运用洛必达法则上下求导就可以了,可能需要运用几次洛必达法则
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x->0
tanx = x+(1/3)x^3+o(x^3)
tan(tanx)
=tan[x+(1/3)x^3+o(x^3)]
= (x+ (1/3)x^3) + (1/3)(x+(1/3)x^3)^3 +o(x^3)
= (x+ (1/3)x^3) + (1/3)(x^3 +o(x^3)) +o(x^3)
= x + (2/3)x^3 +o(x^3)
sinx = x-(1/6)x^3+o(x^3)
sin(sinx)
=sin[x-(1/6)x^3+o(x^3)]
= (x-(1/6)x^3) - (1/6)(x-(1/6)x^3)^3 +o(x^3)
= (x-(1/6)x^3) - (1/6)(x^3+o(x^3)) +o(x^3)
= x - (1/3)x^3 +o(x^3)
tan(tanx) -sin(sinx)
=[ x + (2/3)x^3 +o(x^3) ] -[ x - (1/3)x^3 +o(x^3)]
=x^3 +o(x^3)
lim(x->0) [ tan(tanx) -sin(sinx) ]/x^3
=lim(x->0) x^3/x^3
=1
tanx = x+(1/3)x^3+o(x^3)
tan(tanx)
=tan[x+(1/3)x^3+o(x^3)]
= (x+ (1/3)x^3) + (1/3)(x+(1/3)x^3)^3 +o(x^3)
= (x+ (1/3)x^3) + (1/3)(x^3 +o(x^3)) +o(x^3)
= x + (2/3)x^3 +o(x^3)
sinx = x-(1/6)x^3+o(x^3)
sin(sinx)
=sin[x-(1/6)x^3+o(x^3)]
= (x-(1/6)x^3) - (1/6)(x-(1/6)x^3)^3 +o(x^3)
= (x-(1/6)x^3) - (1/6)(x^3+o(x^3)) +o(x^3)
= x - (1/3)x^3 +o(x^3)
tan(tanx) -sin(sinx)
=[ x + (2/3)x^3 +o(x^3) ] -[ x - (1/3)x^3 +o(x^3)]
=x^3 +o(x^3)
lim(x->0) [ tan(tanx) -sin(sinx) ]/x^3
=lim(x->0) x^3/x^3
=1
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