Answer to Question #95330 in Real Analysis for Rajni

Question

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?

Answer 1

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 } }" $=\lim _{ x\rightarrow 0 }{ \left( 7{ x }^{ 5 }-{ x }^{ 4 }\cos { \frac { 1 }{ x } } \right) } =0$