Limit of a Ratio of Two Functions

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$:

$ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} =  \frac{A}{B} = \frac{\lim_{x \rightarrow a} f(x) }{\lim_{x \rightarrow a} g(x) } $.

$\square$

References: