tan5xtan3x+x趋于兀2

乐畏盲4783求sin3x/tan5x的极限题目:lim(x→π)sin3x/tan5x答案是: - 3/5这样做为什么错:sin3x~3x,tan5x~5x,原式等=3x/5x=3/5 ?请大家给我讲明白,谢谢了. -
弓残琳18854247742 ______[答案] sin3x~3x tan5x~5x 条件是x→0 而本题条件是x→π sinx=-sin3(x-π) tan5x =tan5(x-π) 此时 sin3(x-π)~3(x-π) tan5(x-π)~(x-π lim(x→π)sin3x/tan5x =lim(x→π)-sin3(x-π)/tan5(x-π) =lim(x→π)-3(x-π)/5(x-π) =-3/5

乐畏盲4783已知tanx tany 是方程x²+3√3x+4=0的两根,且x y∈(﹣π/2,π/2),求x+y的值
弓残琳18854247742 ______ tanx +tany=﹣3√3 ,tanxtany=4 ,tan(x+y)=(tanx+tany)/1-tanxtany=√3,由tanx +tany=﹣3√3 ,tanxtany=4 和x y∈(﹣π/2,π/2)可知道x y∈(﹣π/2,0),x+y∈(﹣π,0) 得到x+y=5/3π 应该够详细了 不懂再问哈 呵呵

乐畏盲4783设y=tan(5+3x),求y' -
弓残琳18854247742 ______ y'=[(5+3x)/cos²x]3=(15+9x)/cos²x

乐畏盲4783limtan5(x - 1)/3x - 3(x趋于1)求极限 -
弓残琳18854247742 ______[答案] lim(x->1)tan5(x-1)/(3x-3) let y = x-1 lim(x->1)tan5(x-1)/(3x-3) =lim(x->1)tan5(x-1)/(3x-3) =lim(y->0)tan5y/(3y) =lim(y->0)5y/(3y) =5/3

乐畏盲4783y=tan(x+5),x≠3丌/10+k丌,﹙k∈Z﹚是奇函数还是偶函数? -
弓残琳18854247742 ______ (1)y=tanx的单调周期为,x∈(kπ-π/2,kπ+π/2),在区间上为单调增函数 所以tan(x+π/5)中,kπ-π/2

乐畏盲4783当x趋近于pai lim[sin(3x)/tan(5x)] -
弓残琳18854247742 ______[答案] lim【x→π】[sin(3x)]/[tan(5x)] =lim【x→π】[-sin(3x-3π)]/[tan(5x-5π)] =lim【x→π】-(3x-3π)/(5x-5π) 【等价无穷小代换】 =lim【x→π】-3/5 【上式洛必达法则得到的】 =-3/5

乐畏盲4783求证:tan5x+tan3xcos2x•cos4x=4(tan5x−tan3x). -
弓残琳18854247742 ______[答案] 证明:要证tan5x+tan3xcos2x•cos4x=4(tan5x−tan3x)成立即证sin5xcos5x+sin3xcos3xcos2xcos4x=4(sin5xcos5x−sin3xcos3x)成立,即sin5xcos3x+cos5xsin3xcos2xcos4x=4(sin5xcos3x−cos5xsin3x)即证sin8xcos2xcos4...

乐畏盲4783sin3x/tan5x(当x趋向于派时)刚学啊,似懂非懂,能不能给个详细步骤啊,亲 -
弓残琳18854247742 ______[答案] lim (x-π) sin3x/tan5x =lim (x-π) -sin(3π-3x)/tan(5π-5x) =lim(t→0)-sin3t/tan5t =-3/5 换元,等价无穷小

乐畏盲4783求证tanxtan2x+tan2xtan3x+tan3xtan4x+.+tan(m - 1)xtanmx=tanmx/tanx - m -
弓残琳18854247742 ______[答案] tanx=[tanmx-tan(m-1)x]/{1+} 根据差角的正切公式1+tan(m-1)xtanmx=[tanmx-tan(m-1)x]/tanxtan(m-1)xtanmx=-1+[tanmx-tan(m-1)x]/tanx把每一项都表示成为两项之差,刚好相邻的正负可以抵消tanxtan2x+tan2xtan3x+tan3x...

乐畏盲4783求函数y=tan(3x+3分之派的定义域) -
弓残琳18854247742 ______[答案] 由y=tan(x)函数可知x的定义域是(-π/2+kπ,π/2+kπ)k∈z 所以-π/2+kπ解不等式得:-5π/18所以定义域为(-5π/18+kπ/3,π/18+kπ/3)k∈z 希望能够帮到你哦!