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If $f(x)=\left\{\begin{array}{ll}x\left(1+\frac{1}{2} \sin \left(\log x^2\right)\right), & x \neq 0 \\ 0, & x=0\end{array}\right.$, then
$\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}$
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$\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}$
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Verified Answer
The correct answer is:
does not exist
$\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{x\left(1+\frac{1}{2} \sin \left(\log x^2\right)\right)-0}{x}$
$=\lim _{x \rightarrow 0} 1+\frac{1}{2} \sin \left(\log x^2\right)=1+\frac{1}{2} \sin \left(\lim _{x \rightarrow 0} \log x^2\right)$
$\because \quad \lim _{x \rightarrow 0} \log x^2 \rightarrow-\infty \Rightarrow \sin \left(\lim _{x \rightarrow 0} \log x^2\right)$
is an oscillatory value between 1 and -1 .
$\therefore \quad \lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}$ does not exist.
$=\lim _{x \rightarrow 0} 1+\frac{1}{2} \sin \left(\log x^2\right)=1+\frac{1}{2} \sin \left(\lim _{x \rightarrow 0} \log x^2\right)$
$\because \quad \lim _{x \rightarrow 0} \log x^2 \rightarrow-\infty \Rightarrow \sin \left(\lim _{x \rightarrow 0} \log x^2\right)$
is an oscillatory value between 1 and -1 .
$\therefore \quad \lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}$ does not exist.
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