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If $$\lim_{x \rightarrow a}\frac{f(x)}{g(x)}=0$$ where $a$ is a real number, or could be $ \pm\infty$, and $f,g$ are continuous functions (not necessary polynomials), with $g(a)\neq 0$.

what we can conclude about $f$ and $g$ ?

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Rational function usually means that $f$ and $g$ are polynomials. Please clarify. –  lhf Mar 22 '11 at 1:14
    
I fixed it, thanks. –  Annoli Mar 22 '11 at 1:17

3 Answers 3

Edit: I had misread part of the question, and I've fixed my answer accordingly.

In the real number case, if $g(a)\neq 0$, then the function $h(x)=\frac{f(x)}{g(x)}$ is defined and continuous near $x=a$, and the limit is $h(a)=\frac{f(a)}{g(a)}$. The limit is $0$ if and only if $f(a)=0$.

If $g(a)=0$, then $h(x)$ will be defined near $a$ as long as $g(x)\neq 0$ for all $x\neq a$ in some interval containing $a$. In that case, $\lim_{x\to a}h(x)=0$ means that for all $\varepsilon>0$ there exists $\delta>0$ such that $f(x)\leq \varepsilon\cdot g(x)$ whenever $0<|x-a|<\delta$.

If $a=\pm\infty$ instead of a real number, say $a=\infty$, then by $g(a)\neq 0$ perhaps you mean $\lim_{x\to\infty}g(x)$ exists and is a nonzero real number. In that case, $\lim_{x\to\infty}h(x)$ exists if and only if $\lim_{x\to\infty}f(x)$ exists, and in that case $\lim_{x\to\infty}h(x)=\frac{\lim_{x\to\infty}f(x)}{\lim_{x\to\infty}g(x)}$. In particular, the limit would be $0$ if and only if $\lim_{x\to\infty}f(x)=0$. For the general case with $a=\infty$, see PEV's answer.

These are essentially just the definitions, and I don't know what else you'd want in general.

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If we have the following: $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$$ then $f = o(g(x))$. In other words, for every $M >0$ there exists a constant $x_0$ such that $|f(x)| \leq M|g(x)|$ for all $x > x_0$. Basically this means that $g(x)$ grows faster than $f(x)$. This is in contrast with big-O notation where the inequality only has to hold for one $M$.

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That was very helpful. But what about $-\infty$? –  Annoli Mar 22 '11 at 1:35
    
@Helen: It would be the same except with $x<x_0$ instead of $x>x_0$. –  Jonas Meyer Mar 22 '11 at 1:38

Remember the limit laws:

  • If $\lim\limits_{x\to a}\;h(x)$ and $\lim\limits_{x\to a}\;\ell(x)$ both exists, and $\lim\limits_{x\to a}\'\ell(x)\neq 0$, then $$\lim_{x\to a}\frac{h(x)}{\ell(x)} = \frac{\lim\limits_{x\to a}\;h(x)}{\lim\limits_{x\to a}\;\ell(x)}.$$ That is, the limit of a quotient is the quotient of the limits, provided they both exist, and the limit in the denominator is not equal to $0$.

This is also true when we take one-sided limits, or limits as $x\to\infty$ or $x\to-\infty$.

In particular, if $$\lim_{x\to a}\frac{f(x)}{g(x)}\text{ and }\lim_{x\to a}\;g(x)$$ both exist, then $$\lim_{x\to a}\;f(x) = \lim_{x\to a}\left(\frac{f(x)}{g(x)}g(x)\right) = \left(\lim_{x\to a}\frac{f(x)}{g(x)}\right)\left(\lim_{x\to a}\;g(x)\right).$$ This holds, provided of course that $g(x)\neq 0$ in a neighborhood of $a$ (so that $\frac{f(x)}{g(x)}$ is defined on a neighborhood of $a$).

Here we are assuming that $g$ is continuous at $a$ and $g(a)\neq 0$. That means that:

  1. $\lim\limits_{x\to a}\;g(x)$ exists and equals $g(a)$; and
  2. There is an open neighborhood of $a$ where $g(x)\neq 0$.

So we can apply the rule above to conclude that $$\lim_{x\to a}\;f(x) = \left(\lim_{x\to a}\frac{f(x)}{g(x)}\right)\left(\lim_{x\to a}\;g(x)\right) = 0g(a) = 0.$$

This also holds for one sided limits (assuming only that $g$ is continuous from that side); or limits as $x\to\infty$ (assuming then that $\lim\limits_{x\to\infty}\;g(x)$ exists and is nonzero) or $x\to-\infty$ (under the corresponding assumptions). We can even do it dropping the assumption that $g(x)$ is continuous at $a$, requiring only that the limit exist and be nonzero.

If we are also assuming that $f(x)$ is continuous at $a$, then that means that $$f(a) = \lim_{x\to a}\;f(x) = 0.$$

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