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Question

# For x∈R, let f(x)=⎧⎨⎩x3sin(1x),x≠00,x=0. Then which of the following options is (are) TRUE?

A
f is differentiable at x=0
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B
limx0f(x)x2=0
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C
f(x)x3 is continuous at x=0
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D
limxf(x)x2=1
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Solution

## The correct options are A f is differentiable at x=0 B limx→0f(x)x2=0 D limx→∞f(x)x2=1limx→0f(x)=limx→0x3sin(1x) =0 [By sandwich theorem of limits] =f(0) Hence, f is continuous at x=0. Checking differentiability at x=0: f′(0)=limx→0f(x)−f(0)x−0 =limx→0x3sin(1x)x =limx→0x2sin(1x) =0 Hence, f is differentiable at x=0. limx→0f(x)x2 =limx→0xsin(1x) =0 [By sandwich theorem of limits] limx→0f(x)x3 =limx→0sin(1x) which does not exist. Hence, f(x)x3 is not continuous at x=0. limx→∞f(x)x2 =limx→∞xsin(1x) =limh→0sinhh [Putting x=1h] =1

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