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Given two differentiable functions $f$ and $g$ from $\mathbb{R}$ to $\mathbb{R}$ such that $\lim_{x \to a} \frac{f(x)}{g(x)}=L$, where $a$ and $L$ may be any real number or $\infty$.

Question:

Is it necessary that $\lim_{x \to a} f(x)= L \cdot \lim _{x \to a} g(x)$?

Luiz Cordeiro
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Mathslover
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1 Answers1

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There are a couple main reasons why this need not follow.

  1. $\lim_{x\to a} f(x)$ need not exist! Consider something like $a=\infty$, $f(x)=g(x)=\sin(x)+2$ (the $+2$ is so that $g(x)$ isn't $0$). Then $L=1$ but both of the other limits diverge due to oscillation. Since $f(x)$ and $g(x)$ are differentiable, they're continuous at every real $a$, so you can't have counterexamples where $a$ is real and the limits of $f$ and $g$ don't exist.
  2. As drhab and Shine point out, if $L=\infty$, then multiplying by $L$ doesn't really make sense. Even if you pick a convention like $0*\infty=0$, you're not going to end up with the right answer all the time. If $f(x)=17$ and $g(x)=\begin{cases} x^2 & \text{ if }x\ne0\\5 & \text{ if }x=0 \end{cases}$ and $a=0$, we would need $17=\infty*0$ (this sort of thing is why $\infty*0$ is indeterminate.

Now, a standard calculus fact says that if both $\lim_{x\to a}f(x)$, $\lim_{x\to a}g(x)$ exist, and the latter is not zero, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$. Then you can just multiply both sides by $\lim_{x\to a}g(x)$ to get your result.

Mark S.
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