Answer to Question #95330 in Real Analysis for Rajni

Question #95330
Let f:R --> R be defined as f(x) ={ x^7sin(1/x) if x doesn't equal to 0 , 0 if x=0
Show that f"(0) exists and is equal to zero?
1
Expert's answer
2019-09-30T11:01:38-0400

ANSWER If "f(x)=\\begin{cases} { x }^{ 7 }\\sin { \\frac { 1 }{ x } } ,\\quad if\\quad x\\in \\left( -\\infty ,0 \\right) \\cup \\left( 0,+\\infty \\right) \\quad \\\\ 0\\quad \\quad \\quad \\quad \\quad \\quad ,\\quad if\\quad x=0\\quad \\quad \\end{cases}" then "f'(x)=\\begin{cases} 7{ x }^{ 6 }-\\frac { 1 }{ { x }^{ 2 } } \\cdot { x }^{ 7 }\\cdot \\cos { \\frac { 1 }{ x } } ,\\quad \\quad \\quad if\\quad x\\in \\left( -\\infty ,0 \\right) \\cup \\left( 0,+\\infty \\right) \\quad \\\\ \\quad 0\\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad ,\\quad if\\quad x=0 \\end{cases}" . By the definition "f'(0)=\\lim _{ x\\rightarrow 0 }{ \\frac { f(x)-f(0) }{ x } } =\\lim _{ x\\rightarrow 0 }{ \\frac { { x }^{ 7 }\\sin { \\frac { 1 }{ x } } }{ x } } =0" , "f"(x)=\\lim _{ x\\rightarrow 0 }{ \\frac { f'(x)-f'(0) }{ x } } =\\lim _{ x\\rightarrow 0 }{ \\frac { 7{ x }^{ 6 }-\\quad \\quad { x }^{ 5 }\\cdot \\cos { \\frac { 1 }{ x } } }{ x } }" "=\\lim _{ x\\rightarrow 0 }{ \\left( 7{ x }^{ 5 }-{ x }^{ 4 }\\cos { \\frac { 1 }{ x } } \\right) } =0"


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