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I am having trouble understanding this question on limits. Suppose that $r(x)$ is a function where

$$ \lim_{x\rightarrow 0} \dfrac{r(x)}{x^2} =0 \ . $$

Can someone please explain how, from the first limit I can show that:

$$ \lim_{x\rightarrow 0} \dfrac{r(x)}{x} =0 \ . $$

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    Hint: let s(x)=r(x)/x and t(x)=r(x)/x^2. Translate the hypothesis and the conclusion in terms of the functions s and t. Then express s(x) in terms of t(x). Then...2011-08-11
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    Let $\frac{r(x)}{x^2} = h(x)$, for $x \neq 0$. You are given that $\displaystyle \lim_{x \rightarrow 0} \text{ } h(x) = 0$. Now $\frac{r(x)}{x} = x h(x)$, for $x \neq 0$. Now what can be said about $\displaystyle \lim_{x \rightarrow 0} \text{ } xh(x)$?2011-08-11
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    You've been given proofs below. For an "intuitive" reason: the fact that $\frac{r(x)}{x^2}$ goes to $0$ as $x\to 0$ means that $r(x)$ is going to $0$ "a lot faster" than $x^2$; but $x^2$ itself goes to $0$ "faster" than $x$, so $r(x)$ must be going to $0$ faster than $x$ as well, suggesting the second limit.2011-08-11

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