lim(x~无穷)xsin1/x-1/xsinx
lim(x~无穷)xsin1/x-1/xsinx
求极限lim x→0 (1/xsinx+xsin1/x)
:lim(xsin1/x+1/xsinx)x趋于0
求极限公式,lim x趋于无穷,sinx/x x/sin1/x lim x趋于0,xsin1/x 1/xsinx xsi
证明:Lim (1/xsinx-xsin1/x) =1 x→0
lim→0(xsin1/x+1/xsinx)用夹逼定理x/sinx=1解这条题目,请问这题的极限是多少?
lim x→0 1-cosx/xsinx
lim(x→0)(1-cos2x)/xsinx
X趋向0 lim(xsinx)/(1-cosx)
lim(x→0)(1-cos4x)/xsinx
求解函数极限,lim x趋于无穷 (xsinx)/(根号下x四次方+x二次方+1)=
limx趋近于0时(xsin1/x+1/xsinx)的极限值