求lim【h→0】1/h∫【x-h→x+h】cost^2dt(h>0)

求lim【h→0】1/h∫【x-h→x+h】cost^2dt(h>0) lim(h→0)[(x+h)^3-x^3]/h f(x)在x_0处可导,求lim h→0 f(x_0+h)-f(x_0-h)/5h 的值 1.lim x→0 3x/(sinx-x)2.lim x→0 (1-cosmx)/x^23.lim x→0 sin4x/[(√x+2)-√2]4.lim h →0 [sin(x+h)-sin(x-h)]/h lim {(x+h)^3-x^3}/h h→0 极限 lim[（x+h)^3-x^3]/h当h→0时 lim(h→0)[f(x-h)-f(x)]/h=A中A表示什么 设函数f(x)在点x处可导,试求h→0 lim f(x+h)-f(x-h)/2h的值 lim(h→0) [a^(x+h)+a^(x-h)-2a^x]/h^2有谁帮个忙谢谢 lim h趋向0 (x+h)^2-x^2/h f(x)在x=a处可导, lim(h→0) [f(a+h)-f(a-2h)]/h= 导数的乘法法则推倒uv)'=lim(h→0)[u(x+h)v(x+h)-uv]/h=lim(h→0)[u(x+h)v(x+h)+u(x+h)v-u(x+h)v-uv]/h=lim(h→0)[u(x+h)]×[v(x+h)-v(x)]/h+lim(h→0)[v(x)]×[u(x+h)-u(x)]/h=u(x)v'(x)+u'(x)v(x)=u'v+uv'请问这个第一步lim(h→0)[u(x+h)v(x+h)- 求解一道高数极限题lim[tan(x+h)-2tanx+tan(x-h)]/h² h→0 答案为2sec²xtanx求细节 求极限lim（x+h）平方-x平方/h （h趋于0） 设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值 设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值 设f(x)在x=2处可导,且f'(2)=1,则lim h→0 [ f(2+h)-f(2-h)]/h等于多少, 若函数f(x)在点x=a处可导,则lim(h→0)[（f(x)-f(x+3h))/h等于（）,求过程