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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... –  Did Aug 11 '11 at 12:42
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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 $\displaystyle \lim_{x \rightarrow 0} \text{ } xh(x)$? –  user17762 Aug 11 '11 at 13:11
    
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. –  Arturo Magidin Aug 11 '11 at 18:27

3 Answers 3

Since

$$\lim_{x\rightarrow 0}\frac{r(x)}{x^{2}}=\lim_{x\rightarrow 0}\frac{% r(x)}{x}\cdot \lim_{x\rightarrow 0}\frac{1}{x}=0$$

and

$$\lim_{x\rightarrow 0}\frac{1}{x}\neq 0,$$

we must have

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

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wouldn't it be a $0 \cdot \infty$ indeterminate form? Actually, one could argue that since it's not a one-sided limit, it doesn't exist for $\frac{1}{x}$, so we can't say anything about the existence of the product, right? –  Andy Aug 11 '11 at 18:28
    
@Andy: The limit of the product is $0$ by hypothesis, because $r(x)/x^2$ tends to $0$ with $x$. –  Américo Tavares Aug 11 '11 at 18:43
    
alright :) I didn't have any problems anyway, I was just being nitpicky. :) –  Andy Aug 11 '11 at 18:53

$$ \lim_{x\to0}\frac{r(x)}{x} = \lim_{x\to0}\left(\frac{r(x)}{x^2} \cdot x\right) $$ and go on from there.

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Here is a delta-epsilon proof.

Let $\varepsilon > 0$. Since $$ \mathop {\lim }\limits_{x \to 0} \frac{{r(x)}}{{x^2 }} = 0, $$ there exits $\delta > 0$ such that $$ \bigg|\frac{{r(x)}}{{x^2 }}\bigg| = \bigg|\frac{{r(x)}}{{x^2 }} - 0 \bigg| < \varepsilon $$ for any $x$ such that $0 < |x| < \delta$. Clearly, we can assume that $\delta < 1$. Then, $$ \bigg|\frac{{r(x)}}{x} - 0 \bigg| = \bigg|\frac{{r(x)}}{x}\bigg| < |x|\varepsilon < \delta \varepsilon < \varepsilon $$ (for any $x$ such that $0 < |x| < \delta$), and hence $$ \mathop {\lim }\limits_{x \to 0} \frac{{r(x)}}{{x }} = 0. $$

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