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Let $f:\mathbb{R}\to \mathbb{R}$ be a function. Suppose $\lim_{x \to 0} \frac{f(x)}{x} = 0$. Calculate $\lim_{x \to 0} f(x)$.

According to the answer key, $\lim_{x \to 0} f(x) = 0$. I see $f(x)=x^2$ satisfies both limits, but is there a way "construct" an argument to prove this? I mean, how do I arrive at the answer?

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  • $\begingroup$ If $f(x) \not \to 0$ then $\frac {f(x)}x \to 0$ would be impossible. $\endgroup$
    – fleablood
    Commented Jul 7, 2020 at 23:47
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    $\begingroup$ If $\frac{f(x)}{x}\to 0$, then for all $\varepsilon>0$, there exists a neighborhood of $x=0$ for which $$-\varepsilon x<f(x)<\varepsilon x$$ $\endgroup$
    – Mark Viola
    Commented Jul 8, 2020 at 3:03

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$\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac {f(x)} x x =\lim_{x \to 0}\frac {f(x)} x \lim_{x \to 0}x=(0)(0)=0$.

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$\lim_{x\to 0}\frac {f(x)}x = 0$

For any $\epsilon > 0$ then there exists a $\delta>0$ so that if $|x| < \delta$ then $|\frac{f(x)}x| < \epsilon$. But if $D = \min (\delta, 1)$ then if $|x| < D$ then $|f(x)|=|\frac {f(x)}1|< |\frac {f(x)}x| < \epsilon$.

So $\lim_{x\to 0}f(x) = 0$.

Note: If $\lim_{x\to 0} f(x) =k\ne 0$ or if $\lim_{x\to 0} f(x) = \pm \infty$ then $\lim_{x\to 0} \frac {f(x)}{x} = \frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} x} $ is not defined and can not be $0$. But this allows the possibility the $\lim f(x) $ simply does not exist.

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