Quotient Rule of Limit: Proof [ε-δ Method] | Reciprocal Rule of Limit

The quotient rule of limit says that the limit of the quotient of two functions is the same as the quotient of the limit of the individual functions. In this post, we will prove the quotient law of limit by the epsilon-delta method.

Question: What is an epsilon-delta proof of the division rule of limit? Here, you will find the answer given below.
Let’s recall the epsilon-delta definition of limit: Let $f(x)$ be a function such that $\lim\limits_{x \to a} f(x) = A$. Then for every $\epsilon>0$, there exists a $\delta>0$ such that whenever $0<|x-a|<\delta$ we have that

$|f(x)-A| < \epsilon$ …(I)

Using this definition, we will now prove the division rule of limit.

Quotient Law of limit

If the limit of two functions f and g exist when x tends to a, then
$\lim\limits_{x \to a}(\dfrac{f}{g})$ $=\dfrac{\lim\limits_{x \to a} f}{\lim\limits_{x \to a} g}$
Let $\epsilon>0$ be an arbitrary given positive number.
To prove the quotient rule of limit, we will first prove the reciprocal law of limit which says that:
$\lim\limits_{x \to a}(\dfrac{1}{f(x)})$ $=\dfrac{1}{\lim\limits_{x \to a} f(x)}$.
Let $\lim\limits_{x \to a} f(x) = A$.
Then by (I), for $|A|/2>0$, there is a $\delta_1>0$ such that whenever $0<|x-a|<\delta_1$ we have that
$|f(x)-A|<|A|/2$ …(II)
Now, for $0<|x-a|<\delta_1$ we have
$|A| = |A-f(x)+f(x)|$
$\leq |A-f(x)|+|f(x)|$ by the triangle inequality
$< |A|/2+|f(x)|$ by (II)
Thus, $|A| < |A|/2+|f(x)|$.
This implies that $|A|/2 < |f(x)|$
$\Rightarrow 1/|f(x)| < 2/|A|$ …(III)
Now, for for $\dfrac{|A|^2}{2} \epsilon>0$, there is a $\delta_2>0$ such that whenever $0<|x-a|<\delta_2$ we have that
$|f(x)-A|<\dfrac{|A|^2}{2} \epsilon$ …(IV)
Set $\delta:=\{\delta_1, \delta_2\}$. Then for $0<|x-a|<\delta$ we have that
$|\dfrac{1}{f(x)}-\dfrac{1}{A}|$ $=|\dfrac{A-f(x)}{Af(x)}|$
$=\dfrac{1}{|A| |f(x)|}|A-f(x)|$
$< \dfrac{1}{|A|} \dfrac{2}{|A|} \dfrac{|A|^2}{2} \epsilon$ by (III) and (IV).
$< \epsilon$
This proves that $\lim\limits_{x \to a}(1/f(x))$ $=\dfrac{1}{\lim\limits_{x \to a} f(x)}$.
Thus, if  $\lim\limits_{x \to a} g(x) = B$ then by the above reciprocal law of limit we have that $\lim\limits_{x \to a}(1/g(x))$ $=\dfrac{1}{B}$.
$\lim\limits_{x \to a}(f(x)/g(x))$
$=\lim\limits_{x \to a}(f(x) \cdot 1/g(x))$
$=\lim\limits_{x \to a}f(x) \cdot \lim\limits_{x \to a}1/g(x)$ by the Product rule of limit: Proof
$=A \cdot 1/B$
Therefore, $\lim\limits_{x \to a}(f/g)$ $=A/B$ $=\dfrac{\lim\limits_{x \to a} f}{\lim\limits_{x \to a} g}$. This completes the proof of the quotient rule of limit.


We know that $\lim\limits_{x\to 0} (x+1)=0$ and $\lim\limits_{x\to 0} (x^2+1)=1$.
So by the above quotient rule of limit, the limit of $\dfrac{x+1}{x^2+1}$ when x tends to 0 is equal \to
$\lim\limits_{x\to 0} \dfrac{x+1}{x^2+1}$
= $\dfrac{\lim\limits_{x\to 0} (x+1)}{\lim\limits_{x \to 0}(x^2+1)}$
= 1/1
= 1
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