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.
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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.
