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Why is it that the following limit is defined if $\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0$?

$$\lim_{x\to a}\frac{f(x)}{g(x)}$$

In contrast, the limit isn't defined if $\lim_{x\to a}f(x) \neq 0$ but $\lim_{x\to a}g(x) = 0$. Why this discrepancy?

Using the epsilon delta definition, I can prove the following:

$$\lim_{x\to a}\frac{1}{g(x)} = \frac{1}{\lim_{x\to a}g(x)}\;\text{provided that}\;\lim_{x\to a}g(x) \neq 0$$

I can also prove the following:

$$\lim_{x\to a}f(x) \cdot h(x) = \lim_{x\to a}f(x) \cdot \lim_{x\to a}h(x)$$

Therefore , taking $h(x) = 1/g(x)$ :

$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\left[f(x) \cdot \frac{1}{g(x)}\right] = \lim_{x\to a}f(x) \cdot \lim_{x\to a}\frac{1}{g(x)}$$

Now, provided that $\lim_{x\to a}g(x) \neq 0$, we can deduce/evaluate the following:

$$\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$$

So how come this limit is defined if $\lim_{x\to a}g(x) = 0$ provided that $\lim_{x\to a}f(x) = 0$? This doesn't make any sense to me.

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  • $\begingroup$ You can imagine comparing their ‘’velocity’‘ as they approach $0$. $\endgroup$ Commented Jun 22, 2024 at 6:27
  • $\begingroup$ @BoweiTang yes I have somewhat of an intuition. But I'm asking for a proof. Why is $\lim_{x\to a}\frac{f(x)}{g(x)}$ defined in some cases when the numerator's limit is also zero but not in other cases when the numerator's limit isn't zero? $\endgroup$ Commented Jun 22, 2024 at 6:30
  • $\begingroup$ Have you learned Taylor's expansion? That works. $\endgroup$ Commented Jun 22, 2024 at 6:34
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    $\begingroup$ That first statement is not true. The limit is sometimes defined and sometimes isn't. See en.wikipedia.org/wiki/Indeterminate_form for details. $\endgroup$ Commented Jun 22, 2024 at 6:51
  • $\begingroup$ @QiaochuYuan thanks that clarifies some doubts. However, I still wonder why 0/0 is categorized as an indeterminate form but not x/0 where $x\neq 0$? Can we be 100% sure that $lim_{x\to a}\frac{f(x)}{g(x)}$ is undefined if $\lim_{x\to a}g(x) = 0$ and $\lim_{x\to a}f(x)\neq 0$? Why? $\endgroup$ Commented Jun 22, 2024 at 7:09

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The problem with $\lim_{x \to a} \frac{f(x)}{g(x)}$, when $\lim_{x \to a} f(x) \neq 0$ and $\lim_{x \to a} g(x) = 0$ is that the expression $\left|\frac{f(x)}{g(x)}\right|$ can obtain aribitrarily large values, which means that if the first limit exists (because in some cases it can exist), it cannot be finite and is either $+\infty$ or $-\infty$.

The proof

Consider $f(x)$ and $g(x)$, such that $\lim_{x\to a} f(x)$ exists, is finite and not equal to $0$, and $\lim_{x \to a } g(x) = 0$, but there is some neighbourhood of $a$ such that $g(x) \neq 0$ in that neighbourhood. Therefore we have $\lim_{x\to a} |f(x)| = A$ for some $A > 0$ and $\lim_{x \to a } |g(x)| = 0$

This means that, we can choose some $\delta_1$, such that: $$|x-a| < \delta_1 \implies ||f(x)| - A| < A \implies |f(x)| > 0$$ For arbitrary $\varepsilon > 0$ we can choose $\delta_2$, such that: $$|x-a| < \delta_2 \implies |g(x)| < \varepsilon \implies \frac{1}{|g(x)|} > \frac{1}{\varepsilon}$$

Now taking $\delta = \min\{\delta_1, \delta_2\}$: $$|x-a| < \delta \implies \frac{|f(x)|}{|g(x)|} > \frac{1}{\varepsilon}$$ but $\frac{1}{\varepsilon}$ can be any positive real number, so the expression $\left|\frac{f(x)}{g(x)}\right|$ can be arbitrarily large.

The examples

Here are $3$ examples of functions of mentioned form with different limit behaviours:

$$\lim_{x \to 0} \frac{\cos(x)}{x^2} = +\infty$$ $$\lim_{x \to 0} \frac{1}{-x^2} = -\infty$$ $$\lim_{x \to 0} \frac{1}{x} \quad \text{Does not exist}$$

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