"f(x)=x^2sin(\\frac1x)"
"f(0)=0;f(1)=sin(1)"
In the interval (0,1):
"0\\leq sin(\\frac1x)\\leq 1"
"0\\leq x^2sin(\\frac1x)\\leq x^2"
So,the function doesn't blow up to infinity,thus, f(x) is bounded.
(b)"f(x)=\\sqrt{x}sin x"
"f(0)=0;f(1)=sin(1)"
"0\\leq sinx \\leq1"
"0\\leq \\sqrt{x}sinx \\leq\\sqrt x"
So,the function doesn't blow up to infinity,thus, f(x) is bounded.
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Thank you for your help assignment expert
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