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If $$\lim\limits_{x \to 1}{\frac{f(x^3) - 3f(x)}{x^3- 3x}} = 10,$$ find $ \lim\limits_{x\to 1}{f(x)}$

I've tried a few things but at the end I couldn't arrive at anything becasue of the $f(x^3)$ which is causing me troubles

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  • $\begingroup$ $\lim\limits_{x\to 1}f(x^3)=\lim\limits_{x\to 1}f(x)=f(1)$ $\endgroup$ – zwim Apr 17 '18 at 22:51
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\begin{align*} \lim_{x\rightarrow 1}\dfrac{f(x^{3})-3f(x)}{x^{3}-3x}&=10\\ \lim_{x\rightarrow 1}\dfrac{f(x^{3})-3f(x)}{x^{3}-3x}\lim_{x\rightarrow 1}(x^{3}-3x)&=10\lim_{x\rightarrow 1}(x^{3}-3x)\\ \lim_{x\rightarrow 1}(f(x^{3})-3f(x))&=10\cdot -2\\ \lim_{x\rightarrow 1}(f(x^{3})-3f(x))&=-20, \end{align*} Now the existence of $\lim_{x\rightarrow 1}f(x)$ implies that $\lim_{x\rightarrow 1}f(x^{3})$ and vice versa, since $x\rightarrow x^{3}$ is strictly increasing on a neighborhood of $x=1$, and we have in which case that $\lim_{x\rightarrow 1}f(x^{3})=\lim_{x\rightarrow 1}f(x)$, so \begin{align*} -2\lim_{x\rightarrow 1}f(x)=-20, \end{align*}
and the limit is $10$.

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HINT: If $f$ is continuous, then $f(1)-3f(1)=10\cdot(1-3)$.

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