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Define $f:\mathbb{R}\longrightarrow \mathbb{R}$ by $$f(x) =\begin{cases} x^{4/3}\cos \left(\frac1x\right) & \text{if } x \neq 0, \\\\ 0 & \text{if } x =0. \end{cases}$$ Prove that $f$ is differentiable at $0$.

I tried taking a two-sided limit test, but I was unable to simplify it. Can someone help me with this problem?

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You may just observe that, for $x \neq0$, as $x \to 0$, $$ \left|\frac{f(x)-f(0)}{x-0}\right|=\left|\frac{f(x)}{x}\right|=\left|x^{1/3}\cos \left(\frac1x\right)\right|\leq\left|x^{1/3}\right|\to 0. $$

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