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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 \ . $

  • 0
    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

3 Answers 3

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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

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

  • 0
    alright :) I didn't have any problems anyway, I was just being nitpicky. :)2011-08-11
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$ \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. $