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Let $f:\mathbb{R}\to \mathbb{R}$ be a function such that $|f(x)|\leq 2|x|$ for every $x \in \mathbb{R}$. Evaluate $\lim_{x\to 0} \frac{f(x^3)}{x}$.

According to the answer key, it is $0$ (which matches mine). I am not so sure about my solution (below) though.

Rewritting $\lim_{x\to 0} \frac{f(x^3)}{x}$ yields

$$\lim_{x\to 0} \frac{f(x^3)}{x} = \lim_{x\to 0} 2\frac{|x^3|}{x} = 2\lim_{x\to 0} \frac{|x^3|}{x}$$

Analyzing one-sided limits, we have:

$$\lim_{x\to 0^-} \frac{-x^3}{x}=\lim_{x\to 0^-} -x^2=0$$ and $$\lim_{x\to 0^+} \frac{x^3}{x}=\lim_{x\to 0^+} x^2=0$$

Both one-sided limits exist and are equal, therefore $$2\lim_{x\to 0} \frac{|x^3|}{x}= 2\cdot0=0$$.

Is my solution correct?

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Subtle points: you should have $$ 0 \leq \left|\lim_{x\to 0}\frac{f(x^3)}{x}\right|\leq \left|\lim_{x\to 0}\frac{2|x^3|}{x}\right|; $$then you can proceed.

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  • $\begingroup$ That's how I should've started, right? Then, $0 \leq \left|\lim_{x\to 0}\frac{f(x^3)}{x}\right|\leq \left|\lim_{x\to 0}\frac{2|x^3|}{x}\right|$ $\rightarrow$ $0 \leq \left|\lim_{x\to 0}\frac{f(x^3)}{x}\right|\leq \left|0|$ $\rightarrow$ $\left|\lim_{x\to 0}\frac{f(x^3)}{x}\right| = 0$. Is that correct? $\endgroup$ Jul 11, 2020 at 0:41
  • $\begingroup$ Correct, by the squeeze theorem $\endgroup$
    – Integrand
    Jul 11, 2020 at 0:43
  • $\begingroup$ Thanks, @Integrand! $\endgroup$ Jul 11, 2020 at 0:46

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