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Consider the function $f \colon\mathbb R \to\mathbb R$ defined by $f(x)= \begin{cases} x^2\sin(1/x); & \text{if }x\ne 0, \\ 0 & \text{if }x=0. \end{cases}$ Use $\varepsilon$-$\delta$ definition to prove that the limit $f'(0)=0$. Now I see that h should equals to delta; and delta should equal to epsilon in this case. Thanks for everyone contributed! |
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Recall the definition of a derivative i.e. $$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}h$$ Hence, we get that $$f'(0) = \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \dfrac{2h^2 \sin(1/h)-0}h = \lim_{h \to 0} 2h \sin(1/h)$$ Now recall that $\vert \sin(y) \vert \leq 1$. Hence, we have that $$\left \vert 2h \sin(1/h) \right \vert \leq \left \vert 2h \right \vert$$ Hence, we have that $$\lim_{h \to 0} \left \vert 2h \sin(1/h) \right \vert \leq \lim_{h \to 0} \left \vert 2h \right \vert = 0$$ Hence, you get that $$f'(0) = 0$$ |
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$$\left|{\dfrac{f(h)-f(0)}{h}}\right|=\left|{\dfrac{2h^2 \sin{\dfrac{1}{h}}}{h}}\right|=2 \left|{h \sin{\dfrac{1}{h}}}\right|<2\left|h\right|<\varepsilon.$$ Choose $\delta<\dfrac{\varepsilon}{2}.$ |
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