In this post, we will at first recall all the properties of the limits, and then will prove them using the epsilon-delta method. The following are some properties of limits.

Table of Contents

## Properties of Limits

Let us consider two functions f(x) and g(x) of the variable x, and let $a$ be a real number. Assuming both the limits of f(x) and g(x) exist when x→a, we have the following list of properties of limits:

**• **The limit of a constant is constant. That is, $\lim\limits_{x \to a}$ c =c, where c is a constant. This is the constant rule of limits.

**• **$\lim\limits_{x \to a}$ [c f(x)] =c $\lim\limits_{x \to a}$ f(x), where c is a constant. This is the constant multiple rule of limits.

**• **$\lim\limits_{x \to a}$ [f(x)+g(x)] = $\lim\limits_{x \to a}$ f(x)+ $\lim\limits_{x \to a}$ g(x). This is the addition rule of limits.

**• **$\lim\limits_{x \to a}$ [f(x)-g(x)] = $\lim\limits_{x \to a}$ f(x)- $\lim\limits_{x \to a}$ g(x). This is the subtraction rule of limits.

**• **$\lim\limits_{x \to a}$ [f(x)⋅g(x)] = $\lim\limits_{x \to a}$ f(x) ⋅ $\lim\limits_{x \to a}$ g(x). This is called the multiplication rule of limits.

**• **$\lim\limits_{x \to a}$ [f(x)/g(x)] = $\dfrac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$, provided that the limit of g(x) when x→a is non-zero. This is called the division rule of limits.

Before going into the proofs, let us recall the definition of the limit of a function.

## Epsilon-Delta Definition of Limit

Let f(x) be a function such that $\lim\limits_{x \to a}$ f(x) = L. This means that for every ε>0, there exists a δ>0 such that

|f(x)-L| < ε,

whenever 0<|x-a|<δ.

### Negation of Epsilon Delta Definition of Limit

Let us now give the negative statement of $\lim\limits_{x \to a}$ f(x) = L. If f(x) does not go to L as x→a, then it is not true that for every ε>0, there exists a δ>0 such that

|f(x)-L| < ε, whenever 0<|x-a|<δ.

So the negation of the ε-δ definition will be as follows:

If f(x) does not go to L as x→a, then there exists some ε>0 such that for every δ>0 ∃ some point x ∈ {x : 0<|x-a|<δ} such that |f(x)-L| ≥ ε holds true.

**Also Read:**

Using this epsilon-delta definition of a function, we will prove the above properties of limits.

## Limit Properties Proof

### Constant Rule of Limit Proof

Using ε-δ definition, prove that $\lim\limits_{x \to a}$ c =c, where c is a constant.

**Proof:**

Let ε>0 be a given positive number. To show $\lim\limits_{x \to a}$ c =c, we need to show that ∃ a δ>0 such that

|f(x)-c| < ε, whenever 0<|x-a|<δ **…(i)**

Notice that as f(x)=c ∀ x, the inequality in **(i)**, that is, |f(x)-c| < ε is true always for any x. Thus, we can choose any δ>0 for which **(i)** will be true. This completes the proof of the constant rule of limit.

### Constant Multiple Rule of Limit Proof

Using ε-δ definition, prove that $\lim\limits_{x \to a}$ [c f(x)] =c $\lim\limits_{x \to a}$ f(x), where c is a constant.

**Proof:**

Let $\lim\limits_{x \to a}$ f(x) = L.

If c=0, then there is nothing to prove.

So assume that c≠0. Note that ε/|c|>0. Thus, by definition we have that for every ε>0, there exists a δ>0 such that

|f(x)-L| < ε/|c|, whenever 0<|x-a|<δ **…(ii)**

Now, whenever 0<|x-a|<δ we have that

|cf(x)-cL| = c|f(x)-L|

< c ⋅ ε/|c| by **(ii)**

= ε

This shows that $\lim\limits_{x \to a}$ [c f(x)] = cL =c $\lim\limits_{x \to a}$ f(x).

### Sum Rule of Limit Proof

Using ε-δ definition, prove the addition rule of limit, that is, prove that $\lim\limits_{x \to a}$ [f(x)+g(x)] =$\lim\limits_{x \to a}$ f(x)+ $\lim\limits_{x \to a}$ g(x).

**Proof:**

Let $\lim\limits_{x \to a}$ f(x) = L and $\lim\limits_{x \to a}$ g(x) = M. Then by the definition of limits, for a given ε>0, there exist a δ_{1}>0 and a δ_{2}>0 such that

|f(x)-L| < ε/2, whenever 0<|x-a|<δ_{1} **…(iii)**

|g(x)-M| < ε/2, whenever 0<|x-a|<δ_{2} **…(iv)**

Choose δ = min {δ_{1} , δ_{2 }}. Then whenever 0<|x-a|<δ, we have that

|f(x)+g(x) – (L+M)| = |(f(x) – L) + (g(x) – M)|

≤ |f(x) – L| + |g(x) – M| by the triangle inequality |a+b| ≤ |a|+|b|

≤ ε/2 + ε/2 by **(iii)** and **(iv)**

= ε

This proves that $\lim\limits_{x \to a}$ [f(x)+g(x)] = L+M = $\lim\limits_{x \to a}$ f(x) + $\lim\limits_{x \to a}$ g(x).

### Difference Rule of Limit Proof

With the setting of the proof of the sum rule of limit, whenever 0<|x-a|<δ, we have that

|f(x)-g(x) – (L-M)| = |(f(x) – L) – (g(x) – M)|

≤ |f(x) – L| + |g(x) – M| by the inequality |a-b| ≤ |a|+|b|

≤ ε/2 + ε/2 by **(iii)** and **(iv)**

= ε

This proves that $\lim\limits_{x \to a}$ [f(x)-g(x)] = L+M = $\lim\limits_{x \to a}$ f(x) – $\lim\limits_{x \to a}$ g(x).

### Product Rule of Limit Proof

Using ε-δ definition, prove the product rule of limit, that is, prove that $\lim\limits_{x \to a}$ [f(x)⋅g(x)] = $\lim\limits_{x \to a}$ f(x) ⋅ $\lim\limits_{x \to a}$ g(x).

**Proof:**

Let $\lim\limits_{x \to a}$ f(x) = L and $\lim\limits_{x \to a}$ g(x) = M. Note that

|f(x)g(x) – LM| = |f(x)g(x) – Mf(x) + Mf(x) – LM|

= |f(x) (g(x) – M) + M (f(x) – L)|

≤ |f(x) (g(x) – M)+ |M (f(x) – L)| by the triangle inequality |a+b| ≤ |a|+|b|

= |f(x)| |g(x)- M| + M|f(x) – L| **…(*)**

< ε (*we need to show this*)

Corresponding to ε=1, there exists a δ_{1}>0 such that

|f(x)-L| < 1, whenever 0<|x-a|<δ_{1} **…(v)**

∴By **(v)**, |f(x)| = |f(x)-L+L| ≤ |f(x)-L| + |L| < 1+|L| **…(vi)**

Since $\lim\limits_{x \to a}$ g(x) = M, corresponding to $\dfrac{\epsilon}{2(1+|L|)}>0$, there exists a δ_{2}>0 such that

|g(x)-M| < $\dfrac{\epsilon}{2(1+|L|)}$, whenever 0<|x-a|<δ_{2} **…(vii)**

Again since $\lim\limits_{x \to a}$ f(x) = L, corresponding to $\dfrac{\epsilon}{2(1+|M|)}>0$, there exists a δ_{3}>0 such that

|f(x)-L| < $\dfrac{\epsilon}{2(1+|M|)}$, whenever 0<|x-a|<δ_{3} **…(viii)**

Choose δ = min {δ_{1} , δ_{2 }, δ_{3}}. Then whenever 0<|x-a|<δ, we have from **(*)** that

|f(x)g(x) – LM| ≤ |f(x)| |g(x)- M| + M|f(x) – L|

< |f(x)| |g(x)- M| + (1+|M|) |f(x) – L|

< (1+|L|) ⋅ $\dfrac{\epsilon}{2(1+|L|)}$ + (1+|M|) ⋅ $\dfrac{\epsilon}{2(1+|M|)}$, by **(vi), (vii)** and **(viii)**

< ε/2 + ε/2

= ε

Thus, we obtain that $\lim\limits_{x \to a}$ [f(x)⋅g(x)] = LM = $\lim\limits_{x \to a}$ f(x) ⋅ $\lim\limits_{x \to a}$ g(x). This completes the proof of the product rule of limits.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.