Properties of limits
The properties of limits are direct consequences of the way we operate with functions. When we add one function to another we do it for each and every point. We are adding up their respective expressions, which results in a third function that, for each point, gives the sum of the values of each function we are adding. This naturally gives the idea that a sum of continuous functions is continuous, because if the limit of both functions exist at a point, then the resulting function is also defined at that point and has a known limit. For example: if the limit of a function is 1 at a point and the limit of the other function is 2 at the same point, the limit of the sum should naturally be 3.
Notice that if the limit of one function does not exist at a point, then adding a function whose limit does exist yields a limit of the sum that does not exist there. I'm only mentioning this because some people could be thinking that a limit that does not exist at a point is a hole. Therefore adding a function with a limit that does exist at that point would "fill" that hole. This reasoning is totally flawed. To understand that plot the graph for the sum of the functions. The graph should make it obvious that the limit does not exist at that point.
For multivariable functions the same properties hold true.
The properties
- [math]\displaystyle{ \lim_{x \ \to \ a} [f(x) \pm g(x)] = \lim_{x \ \to \ a} f(x) \pm \lim_{x \ \to \ a} g(x) = L_1 \pm L_2 }[/math] (The limit of the sum | difference is the sum | difference of the limits)
- [math]\displaystyle{ \lim_{x \ \to \ a} f(x)g(x) = \left(\lim_{x \ \to \ a} f(x)\right) \left(\lim_{x \ \to \ a} g(x)\right) = L_1L_2 }[/math] (The limit of the product is the product of the limits. Swap the second function with [math]\displaystyle{ \frac{1}{g(x)} }[/math] and we have the quotient rule, provided that the second function is not zero at that point.)
- [math]\displaystyle{ \lim_{x \ \to \ a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \ \to \ a} f(x)} = \sqrt[n]{L} }[/math]. (The limit of the nth root is the nth root of the limit, provided that the limit is not negative and [math]\displaystyle{ n }[/math] is even)
- [math]\displaystyle{ \lim_{x \ \to \ a} cf(x) = c\lim_{x \ \to \ a} f(x) = cL }[/math]. (The limit of a function times a constant is the constant times the limit. This property is a linear transformation because when we multiply a function by a constant we don't change the critical points, they remain where they already are. We often forget that to multiply by 1 is the same as to multiply by c and divide by c at the same time, provided that c is not zero.)
- [math]\displaystyle{ \lim_{x \ \to \ a} f(\overbrace{g(x)}^{u}) = \lim_{u \ \to \ g(a)} f(u) }[/math]. (Careful with this! The existence of the limit of [math]\displaystyle{ g(x) }[/math] does not guarantee the existence of the limit of [math]\displaystyle{ f(g(x)) }[/math]. However, if the limif of [math]\displaystyle{ g(x) }[/math] does not exist, this guarantees that the limit of [math]\displaystyle{ f(g(x)) }[/math] does not too.)
Proofs of the properties
Most teachers skip those in class because they take a lot of time to properly explain all the details.
[math]\displaystyle{ (f + g)(x) = f(x) + g(x) }[/math] That's the expression that means sum of functions.
[math]\displaystyle{ \lim_{x \ \to \ a} (f + g)(x) = \lim_{x \ \to \ a} f(x) + \lim_{x \ \to \ a} g(x) }[/math] That's how some teachers explain at class. It's not the proof though.
To prove that the limit of the sum is the sum of the limits we have to use the formal definition of a limit. The proof of the triangle inequality is a pre-requisite to do it. In here please notice that the limit of both functions is defined. Otherwise, if one or both limits don't exist, we can't do the sum!
[math]\displaystyle{ \lim_{x \ \to \ a} f(x) = L }[/math] is the same thing as [math]\displaystyle{ \lim_{x \ \to \ a} f(x) - L = 0 }[/math] (Think about it: if the limit is [math]\displaystyle{ f(a) }[/math], then the distance between the limit and [math]\displaystyle{ f(a) }[/math] should be zero because one is equal to the other)
[math]\displaystyle{ f(x) + g(x) - (L_1 + L_2) = [f(x) - L_1] + [g(x) - L_2] }[/math]
We are already assuming that the limit of both functions exists, then let's assume that [math]\displaystyle{ f(x) \to 0 }[/math] and [math]\displaystyle{ g(x) \to 0 }[/math] as [math]\displaystyle{ a \to 0 }[/math]. Some functions may not have a limit at [math]\displaystyle{ x = 0 }[/math] but we already began by excluding them. If both limits are equal to zero, then we have to prove that the sum of limits is also equal to zero. For every [math]\displaystyle{ \epsilon \gt 0 }[/math] there is a [math]\displaystyle{ \delta \gt 0 }[/math] such that
[math]\displaystyle{ |f(x) + g(x)| \lt \epsilon| }[/math] whenever [math]\displaystyle{ 0 \lt |x - a| \lt \delta }[/math]
Let [math]\displaystyle{ \epsilon }[/math] be given. Since [math]\displaystyle{ f(x) \to a }[/math] as [math]\displaystyle{ x \to a }[/math], there is a [math]\displaystyle{ \delta_1 \gt 0 }[/math] such that
[math]\displaystyle{ |f(x)| \lt \frac{\epsilon}{2} }[/math] whenever [math]\displaystyle{ 0 \lt |x - a| \lt \delta_1 }[/math]
Repeat for [math]\displaystyle{ g(x) }[/math]:
[math]\displaystyle{ |g(x)| \lt \frac{\epsilon}{2} }[/math] whenever [math]\displaystyle{ 0 \lt |x - a| \lt \delta_2 }[/math]
(if you didn't grasp the division by 2. Remember that the formal definition of a limit has the limit bounded by +error and -error. We've just used the property of a modulus)
If we let [math]\displaystyle{ \delta }[/math] denote the smaller of the two numbers [math]\displaystyle{ \delta_1 }[/math] and [math]\displaystyle{ \delta_2 }[/math], then both the previous inequalities are valid if [math]\displaystyle{ 0 \lt |x - a| \lt \delta }[/math] and hence, by the triangle inequality, we can find that
[math]\displaystyle{ |f(x) + g(x)| \leq |f(x)| + |g(x)| \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon }[/math].
The other properties follow a similar reasoning.
Note: I followed Tom Apostol in this proof.
Links for the proofs:
