Function:
f(x)={xeπx−1,x≠01π,x=0
limx→0f(x)=1π=f(0)⇒Function is continuous at x=0
f′(0)=limh→0f(h)−f(0)h=−12
f′(0)=limx→0f(x)−f(0)x
=limx→0x2sin(1/x)x
=limx→0xsin(1x)=0
For x≠0,
f′(x)=2xsin(1x)−cos(1x)
Answer: f′(0)=0 ✅
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