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Let's say we have 2 functions $f(x)$ and $g(x)$.

  • $f(x)$ and $g(x)$ go to zero when $x \to a$.
  • $|f(x)| < |g(x)|$ for every $x$ such as $|x-a| < \epsilon$ for every positive $\epsilon$ ($\epsilon$ is small of course).

Is it always true that $f(x)/g(x) \to 0$ when $x \to a$?

Nikkal
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  • Contrary to your belief, $\epsilon$ does not have to be small (which is still to be defined). The discussion holds for any $\epsilon$, including a googolplex. –  Dec 13 '19 at 19:29
  • You may find this interesting. – J.G. Dec 13 '19 at 19:56

3 Answers3

3

No. Let $f(x)=cx$ and $g(x)=x$, for $-1<c<1$, and $a=0$. Clearly both $f(x)\to0$ and $g(x)\to0$ as $x\to0$. Then $|f(x)|<|g(x)|$ is true for all $x\neq 0$ and $\frac{f(x)}{g(x)}\to c$ as $x\to 0$.

So, $\frac{f(x)}{g(x)}$ can converge to any real number $c\in(-1,1)$ as $x\to 0$.

2

A famous counterexample is $f(x)=\sin x,\ g(x)=x,\ a=0$. Then, $f(a)=g(a)=0$ and $|f(x)|<|g(x)|$ for all nonzero $x$ but $\displaystyle \frac{f(x)}{g(x)}=\frac{\sin x}{x}\to 1$ as $x\to a$ (Question 75130).

Keeping $a=0$ but taking $f=x,\ g=2\left|x\right|$ or $f=x\sin\left(\frac{1}{x}\right),\ g=x$ shows that $\displaystyle \frac{f(x)}{g(x)}$ doesn't even necessarily have a limit at $a$, despite $f(x)$ and $g(x)$ having one.

Jam
  • 10,325
1

If $g(x)=2f(x)$, we do have $|f(x)|<g(x)|$ but the ratio is always $\dfrac12$ !