Proofs of all Limit Properties [ε-δ Definition]

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. 

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 δ1>0 and a δ2>0 such that 

|f(x)-L| < ε/2, whenever 0<|x-a|<δ1 …(iii)

|g(x)-M| < ε/2, whenever 0<|x-a|<δ2 …(iv)

Choose δ = min {δ1 , δ2 }. 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 δ1>0 such that 

|f(x)-L| < 1, whenever 0<|x-a|<δ1 …(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 δ2>0 such that 

|g(x)-M| < $\dfrac{\epsilon}{2(1+|L|)}$, whenever 0<|x-a|<δ2 …(vii)

Again since $\lim\limits_{x \to a}$ f(x) = L, corresponding to $\dfrac{\epsilon}{2(1+|M|)}>0$, there exists a δ3>0 such that 

|f(x)-L| < $\dfrac{\epsilon}{2(1+|M|)}$, whenever 0<|x-a|<δ3 …(viii)

 

Choose δ = min {δ1 , δ2 , δ3}. 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.