If $\;\lim\limits_{x\rightarrow 10}g(x)=a\;$ and if$\;\lim\limits_{x \rightarrow10}\dfrac{f(x)}{g(x)}=b$, find $\lim\limits_{x\rightarrow10}f(x).$

Question

If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$

The answer is insufficient information to determine. But I thought there was a "limit multiplication rule" I could use to determine it's $ab$? Can someone give an example when this doesn't hold?

[EDIT] based on below discussion, I guess the question boils down to, can we conclude $\lim\limits_{x\rightarrow10}f(x)$ exists based on above conditions?

I am also curious if the answer changes if $b=0$ or $a=0$?

Answer 1

$f(x)= \frac{f(x)}{g(x)} \cdot g(x)$ for $x $ in a neighborhood of $10$, since $a \ne 0.$

Hence

$$f(x)= \frac{f(x)}{g(x)} \cdot g(x) \to b \cdot a$$

as $x \to 10.$

Answer 2

If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$

Since $\;a\ne0\;,\;$ there exists $\;\delta>0\;$ such that $\;g(x)\ne0\;$ for any $\;x\in\left]10-\delta,10+\delta\right[\;.$

Consequently,

$f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\quad\forall\;x\in\left]10-\delta,10+\delta\right[\;,$

and, by applying the limit multiplication rule, we get that

$\begin{align} \lim\limits_{x\rightarrow10}f(x)&=\lim\limits_{x\to10}\left(g(x)\cdot\dfrac{f(x)}{g(x)}\right)=\\ &=\lim\limits_{x\to10}g(x)\cdot\lim\limits_{x\to10}\dfrac{f(x)}{g(x)}=a\cdot b\;. \end{align}$

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Now we are going to prove the following property

If there exists $\;\delta>0\;$ such that $\;g(x)\ne0\;$ for any $\;x\in\left]10-\delta,10+\delta\right[\;,\;$ if $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ then $\;\lim\limits_{x\rightarrow10}f(x)=a\cdot b\;.$

Since $\;f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\quad\forall\;x\in\left]10-\delta,10+\delta\right[\;,$

by applying the limit multiplication rule, we get that

$\begin{align} \lim\limits_{x\rightarrow10}f(x)&=\lim\limits_{x\to10}\left(g(x)\cdot\dfrac{f(x)}{g(x)}\right)=\\ &=\lim\limits_{x\to10}g(x)\cdot\lim\limits_{x\to10}\dfrac{f(x)}{g(x)}=a\cdot b\;. \end{align}$