The correct option is C 0
limx→0f(x)x=limx→0f(x)−f(0)x−0=f′(0)
f(x)=∣∣
∣∣cosxx12sinxx22xtanxx1∣∣
∣∣
f′(x)=∣∣
∣∣−sinx102sinxx22xtanxx1∣∣
∣∣+∣∣
∣∣cosxx12cosx2x2tanxx1∣∣
∣∣+∣∣
∣
∣∣cosxx12sinxx22xsec2x10∣∣
∣
∣∣
Thus, limx→0f(x)x=f′(0)
=∣∣
∣∣010000001∣∣
∣∣+∣∣
∣∣101202001∣∣
∣∣+∣∣
∣∣101000110∣∣
∣∣
=0
Alternatively,
f(x)=∣∣
∣∣cosxx12sinxx22xtanxx1∣∣
∣∣
R3→R3−R1, we get
f(x)=∣∣
∣∣cosxx12sinxx22xtanx−cosx00∣∣
∣∣
Expanding along R3,
f(x)=(tanx−cosx)x2
limx→0f(x)x=limx→0(tanx−cosx)x=0