设u＝1x，则当x→∞时，u→0．于是：limx→∞(sin1x+cos1x)x＝limu→0(sinu+cosu)1u=limu→0(1+sinu+cosu?1)1sinu+cosu?1?sinu+cosu?1u=limu→0esinu+cosu?1u＝elimu→0sinu+cosu?1u=e．
lim(x->∞) sin(1/x)cos(1/x)=sin0.cos0 =0
= 2sin(X ^ 2)* [(X-0)/(sinx的-SIN0)]只 = 2sin(X ^ 2)* [ 1 /(sinx的导数)]只 = 2sin(X ^ 2)*(1 / cosx)限制 = 2sin(0 ^ 2)*(1 / COS0)= 0 分析总结。 利用取对数的方法求limx趋于无穷大sin1xcos1xx的极限结果一 题目 利用取对数的方法求limx趋于无穷大(sin1/x+c...
x=lim(sint+cost)^(1/t)=lim[1+(sint+cost-1)]^{[1/(sint+cost-1)]*(sint+cost-1)/t}因为lim(t->0)(sint+cost-1)/t=lim(t->0)(cost-sint)=1所以lim(sint+cost)^(1/t)=lim[1+(sint+cost-1)]^{[1/(sint+cost-1)]*(sint+cost-1)/t}...
解：原式=limt→0(sint+cost)1t=exp[limt→0ln(1+sint+cost−1)t]=exp[limt→0sint+cost−1t]=exp[limt→0sintt+limt→0cost−1t]=exp[1+limt→0(−sint)1]=exp[1+0]=e
令t=1/x，原式变为t趋近于0时，(3+5t^2)/(5t+3t^2)*sint 因为t趋近于0时，sint与t等价，所以 变为，(3+5t^2)/(5t+3t^2)*t=(3+5t^2)/(5+3t)将t=0代入，得极限值为3/5
简单分析一下，答案如图所示
(sin1/x+cos1/x)x=[√2 sin(1/x+π/4)]x又因为sinx等价于x所以sin(1/x+π/4)等价于(1/x+π/4)所以原式等价于lim[√2 (1/x+π/4)]x=lim[(√2×4/π)(π/(4x)+1)]^(π/(4x)×(4/π)又因为lim(1/x+1)x=e所以原式=lim[(√2×4/π)×e]^(4/π)...
lims-|||-→00-|||-1-|||-lim(sinx cosx)-|||-→0-|||-1-|||-sint+cos-1-|||-=lim(1+sin x cosx-1)sinz+cosz-1-|||-0→0-|||-=explim-|||-sinx cosx-1-|||-x→0-|||-=explim-|||-sin c-|||--x2/2-|||-x→0-|||-=exp(1+0)=e 结果...
cos1/x→1sin1/x~1/xlim(x→∞)(sin1/x-cos1/x)^x=lim(x→∞)(1/x-1)^x=-lim(x→∞)(1-1/x)^x=-lim(x→∞)(1+(-1/x)^(-x)*(-1)=-e^(-1)=-1/e 解析看不懂?免费查看同类题视频解析查看解答 更多答案(1) 相似问题 求极限,请高手指导.lim(x→+∞) [(sin1/x + ...