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Question

Let f(x)=∣ ∣cosxx12sinxx22xtanxx1∣ ∣. The value of limx0f(x)x is equal to

A
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C
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Solution

The correct option is C 0
limx0f(x)x=limx0f(x)f(0)x0=f(0)

f(x)=∣ ∣cosxx12sinxx22xtanxx1∣ ∣

f(x)=∣ ∣sinx102sinxx22xtanxx1∣ ∣+∣ ∣cosxx12cosx2x2tanxx1∣ ∣+∣ ∣ ∣cosxx12sinxx22xsec2x10∣ ∣ ∣

Thus, limx0f(x)x=f(0)
=∣ ∣010000001∣ ∣+∣ ∣101202001∣ ∣+∣ ∣101000110∣ ∣
=0

Alternatively,
f(x)=∣ ∣cosxx12sinxx22xtanxx1∣ ∣

R3R3R1, we get
f(x)=∣ ∣cosxx12sinxx22xtanxcosx00∣ ∣

Expanding along R3,
f(x)=(tanxcosx)x2
limx0f(x)x=limx0(tanxcosx)x=0

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