Tính lim x tiến tới 0 của căn bậc hai của 1+2x . căn bậc ba của 1+3x

1+2x.1+3x3.1+4x4−1=1+2x−1+2x+1+2x.1+3x3−1+2x.1+3x3+1+2x.1+3x3.1+4x4−1=1+2x−1+1+2x1+3x3−1+1+2x.1+3x3.1+4x4−1⇒limx→01+2x.1+3x3.1+4x4−1x=limx→01+2x−1x+limx→01+2x.1+3x3−1x+limx→01+2x.1+3x3.1+4x4−1x

Tính 

limx→01+2x−1x=limx→01+2x−11+2x+1x1+2x+1=limx→02xx1+2x+1=limx→021+2x+1=21+1=1

limx→01+2x.1+3x3−1x=limx→01+2x.1+3x3−11+3x32+1+3x3+1x1+3x32+1+3x3+1=limx→01+2x.3xx1+3x32+1+3x3+1=limx→031+2x1+3x32+1+3x3+1=3.11+1+1=1

limx→01+2x.1+3x3.1+4x4−1x=limx→01+2x.1+3x3.1+4x4−11+4x43+1+4x42+1+4x4+1x1+4x43+1+4x42+1+4x4+1=limx→01+2x.1+3x3.4xx1+4x43+1+4x42+1+4x4+1=limx→041+2x.1+3x3.1+4x43+1+4x42+1+4x4+1=4.1.11+1+1+1=1

Vậy limx→01+2x.1+3x3.1+4x4−1x=1+1+1=3

Đáp án cần chọn là: D