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.