If f(x) = begin{cases} x left(1 + frac{1}{2} | vedclass

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(B) We need to evaluate the limit: $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.Given $f(0) = 0$,the expression becomes $\lim_{x \rightarrow 0} \fra

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does not exist

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(B) We need to evaluate the limit: $\lim_{x ightarrow 0} \frac{f(x) - f(0)}{x}$.Given $f(0) = 0$,the expression becomes $\lim_{x ightarrow 0} \frac{x \left(1 + \frac{1}{2} \sin (\log x^2) ight) - 0}{x}$.Simplifying the expression by canceling $x$ (since $x 
eq 0$ as $x ightarrow 0$):$= \lim_{x ightarrow 0} \left(1 + \frac{1}{2} \sin (\log x^2) ight)$.As $x ightarrow 0$,$x^2 ightarrow 0^+$,so $\log x^2 ightarrow -\infty$.The function $\sin(\log x^2)$ oscillates between $-1$ and $1$ as $x ightarrow 0$.Therefore,the limit $\lim_{x ightarrow 0} \sin(\log x^2)$ does not exist.Consequently,$\lim_{x ightarrow 0} \frac{f(x) - f(0)}{x}$ does not exist.

If $f(x) = \begin{cases} x \left(1 + \frac{1}{2} \sin (\log x^2) \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then find the value of $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.

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