To find \( f'(0) \) for the given function \( \displaystyle f(x) = x + x^3 \sin\left(\frac{1}{x}\right) \), we will use the definition of the derivative.
Solution:
Using the definition of the derivative,
\( \displaystyle f'(0) = \lim_{h \to 0} \frac{f(0 + h)\, – f(0)}{h} \\ \displaystyle = \lim_{h \to 0} \frac{f(h) – f(0)}{h} \\ \)Substitute \( f(h) \) into the limit:
\( \displaystyle f'(0) = \lim_{h \to 0} \frac{h + h^3 \sin\left(\frac{1}{h}\right)}{h} \text{, since } f(0) = 0 \\ \displaystyle = \lim_{h \to 0} \frac{h\left(1 + h^2 \sin\left(\frac{1}{h}\right)\right)}{h} \\ \displaystyle = \lim_{h \to 0} \left(1 + h^2 \sin\left(\frac{1}{h}\right)\right) \\ \displaystyle = 1 + \lim_{h \to 0} h^2 \sin\left(\frac{1}{h}\right) \\ = 1 + 0 = 1 \\ \)Therefore, \( \boxed{f'(0) = 1} \)
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