In this post, I show that $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}$ given that $\lim_{x\rightarrow a}f(x)=A$, $\lim_{x \rightarrow a} g(x) = B$, $B \ne 0$ and $g(x) \ne 0$. To do this, I approximately follow the steps in reference [1]. Known: From the the definition of a limit, Whenever $ 0 < |x-a| < \delta $, $ |f(x) – A| < \epsilon_1$ with $\epsilon_1 > 0$. Whenever $ 0 < |x-a| < \delta $, $ |g(x) – B| < \epsilon_2$ with $\epsilon_2 > 0$. Objective: The objective is to directly show that $ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} =  \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)} $. Proof: Consider the expression, $ \bigg| \frac{1}{g(x)} – \frac{1}{B}  \bigg|$. Using algebra, $ \bigg| \frac{1}{g(x)} – \frac{1}{B}  \bigg| = \bigg| \frac{B – g(x)}{g(x)B} \bigg| = \bigg| \big(g(x) – B \big) \frac{-1}{g(x)B} \bigg|$. From this property, $ \bigg| \frac{1}{g(x)} – \frac{1}{B}  \bigg| = \bigg|    g(x) – B  \bigg| \bigg|   \frac{-1}{g(x)B}   \bigg| = \bigg|   g(x) – B  \bigg| \bigg|  \frac{1}{g(x)} \bigg| \bigg| \frac{1}{B}  \bigg| $. Using what is known about $\epsilon_2$, $ \bigg| \frac{1}{g(x)} – \frac{1}{B}  \bigg| < \epsilon_2 \bigg|  \frac{1}{g(x)} \bigg| \bigg| \frac{1}{B}  \bigg| $. Since $g(x) \ne 0$ and $B \ne 0$, $\bigg|  \frac{1}{g(x)} \bigg| > 0$ and $\bigg| \frac{1}{B}  \bigg| > 0$. And since $\epsilon_2 > 0$, $ \bigg| \frac{1}{g(x)} – \frac{1}{B}  \bigg| < \epsilon_2 \bigg|  \frac{1}{g(x)} \bigg| \bigg| \frac{1}{B}  \bigg| > 0$. Define $\epsilon_3 \equiv \epsilon_2 \bigg|  \frac{1}{g(x)} \bigg| \bigg| \frac{1}{B}  \bigg| > 0$. Then $ \bigg| \frac{1}{g(x)} – \frac{1}{B}  \bigg| < \epsilon_3$ whenever $ 0 < |x-a| < \delta $ with $\epsilon_3 > 0$, or $ \lim_{x \rightarrow a} \frac{1}{g(x)} = \frac{1}{B} $. Using this property $ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} f(x) \lim_{x \rightarrow a} \frac{1}{g(x)}  $. Using $ \lim_{x \rightarrow a} \frac{1}{g(x)} = \frac{1}{B} $ and $ \lim_{x \rightarrow a} f(x) = A$: […]