A @mindhunter mathematical proof

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Hello my fellow mathematical Steemians!

This is my first true mathematical post, and writing mathematical symbols on here is not always easy given the Steemit formatting rules! If you have any suggestions as to how I should present my work minus these formatting issues in the future, then please comment below.

Thanks in advance for everyone who has the patience to read and evaluate all this, it is much appreciated.

I will be studying the limits as x->±∞, of function f:|R->|R, where f(x)=[k,x]∫e^(t^2)dt.
Most specifically, proving that lim(x->-∞)f(x)=-∞ and lim(x->+∞)f(x)=+∞
([k,x] is the interval of the real definite integral that I use to define f.)
k is a real number, picked at random, since it doesn't matter which number you choose for this particular proof.

I have divided the proof in steps, since I think it is easier to be read and studied that way.
The steps I take might seem a bit off at times, but hang in there till the end and you shall understand :)

Step 1: Study of f and df/dx, concerning their monotonic behaviour

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e^(x^2) is a continuous function, since it is the product of composition of two continuous functions. x^2 as a polynomial function and e^x as an exponential.

From the above, and taking in consideration the definition of the integral, we conclude that we can define df/dx.
df/dx=f'(x)=e^(x^2), x∈ |R
It is obvious, since x is a real number, that df/dx>0, for any x∈ |R.
So, we have that f is a monotonically increasing function, for every x.

d(df/dx)/dx=f''(x)=2xe^(x^2)
f''(x)=0 <=> 2xe^(x^2)=0 <=> x=0 since 2>0, e^(x^2)>0.
f''(x)>0 <=> x>0 and f''(x)<0 <=> x<0, for the same reasons.
lim(x->0)f''(x)=lim(x->0)2xe^(x^2)=20e^0=f''(0), so f'' is continuous at x=0
From the above we conclude that df/dx is motonically increasing for every x>=0, and monotonically decreasing for every x<=0.

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Source

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Step 2: Making sure the limits we are trying to find really exist!

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We know that f is a continuous, monotonically increasing function.

There is a theorem(which I think was proved by Weierstrass, though I don't remember its name) which requires a real function with the above qualities f has(monotonically incr/decr and continuous), saying that we can define the set of values Rf of the function to be:
Rf=(lim(x->-∞)f(x),lim(x->+∞)f(x)).

The above theorem, though it seems completely useless for this proof, has a secret trait. It does not only help us find the set of values of a function g:[a,b]->|R, it indirectly states that the limits of the function, as x approaches a and b, exist and can be found.

To conclude this step, I'm going to set lim(x->-∞)f(x)=L1 and lim(x->+∞)f(x)=L2, where L1,L2∈ {|R,±∞}.

Step 3: Reductio ad absurdum.

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Every limit can have one of three values: It can exist and be finite, it can exist and be infinite and it can be non-existent.Since we know L1,L2 exist(Step 2), we can use what we call "Reductio ad absurdum".
"Let L1,L2 be real, finite numbers, and not something infinite."

Step 4: Calculating L1 and L2 (finally!!!).

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Let a be a random real positive number.
We will find lim(x->+∞)f(x+a) and lim(x->-∞)f(x+a).
If t=x+a, then if x->+∞, then t->+∞.
So we have that lim(x->+∞)f(x+a)=lim(t->+∞)f(t), which means that lim(x->+∞)f(x+a)=L2.
Using the same technique we can say that lim(x->-∞)f(x+a)=lim(t->-∞)f(t)=L1.

L1

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Considering x to be a random real number for which x<=(-a), we define B=[x,x+a].
Applying the Mean value theorem of Calculus to f(we can do that, check the qualities f is proven to have at the beginning of step 1) for the set B, we have that:
There exists k2∈ B : f'(k2)=(f(x+a) - f(x))/(x+a-x)=(f(x+a) - f(x))/a .
k2∈ B => x < k2 < x+a
Since f' is monotonically decreasing for x<=0, f'(x) > f'(k1) > f'(x+a).
=> af'(x) > f(x+a)-f(x) > af'(x+a) .

lim(x->-∞)af'(x)=lim(x->-∞)ae^(x^2)=+∞ (x->-∞ => x^2->+∞ => ae^(x^2)->+∞ , because a>0)
lim(x->-∞)af'(x+a)=lim(x->-∞)ae^[(x+a)^2]=+∞ (x->-∞ => (x+a)->-∞ => [(x+a)^2]->+∞ => ae^[(x+a)^2]->+∞ , because a>0)
Using the squeeze/sandwich theorem, we have:
lim(x->-∞)[f(x+a)-f(x)]=+∞
Previously, we proved both lim(x->-∞)f(x+a) and lim(x->-∞)f(x) to exist, and using that information, we can say that [lim(x->-∞)f(x)] - [lim(x->-∞)f(x)]=+∞
=>L1-L1=+∞ => 0=+∞ , since L1∈ |R.

The last statement is obviously false. We said that L1∈ |R and using that we concluded to a completely false statement, so the first statement must be false, so L1 is not finite, but rather an infinite element, so L1=±∞ .
In step 2, we showed that Rf=(L1,L2), so L2 can't be equal to minus infinity, therefore it must be L1=-∞ !!!

L2

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Considering x to be a random real positive number, we define A=[x,x+a].

Applying the Mean value theorem of Calculus to f(we can do that, check the qualities f is proven to have at the beginning of step 1) for the set A, we have that:
There exists k1∈ A : f'(k1)=(f(x+a) - f(x))/(x+a-x)=(f(x+a) - f(x))/a .
k1∈ A => x < k1 < x+a
Since f' is monotonically increasing for x>=0, f'(x) < f'(k1) < f'(x+a).
=> af'(x) < f(x+a)-f(x) < af'(x+a) .

lim(x->+∞)af'(x)=lim(x->+∞)ae^(x^2)=+∞
lim(x->+∞)af'(x+a)=lim(x->+∞)ae^[(x+a)^2]=+∞
Using the squeeze/sandwich theorem, we have:
lim(x->+∞)[f(x+a)-f(x)]=+∞

Previously, we proved both lim(x->+∞)f(x+a) and lim(x->+∞)f(x) to exist, and using that information, we can say that [lim(x->+∞)f(x)] - [lim(x->+∞)f(x)]=+∞
=>L2-L2=+∞ => 0=+∞ , since L2∈ |R.

The last statement is obviously false. We said that L2∈ |R and using that we concluded to a completely false statement, so the first statement must be false, so L2 is not finite, but rather an infinite element, so L2=±∞ .
In step 2, we showed that Rf=(L1,L2), so L2 can't be equal to minus infinity, therefore it must be L2=+∞ !!!

We have proved exactly what we wanted!
lim(x->-∞)f(x)=-∞
lim(x->+∞)f(x)=+∞

Mathematics is a way of life for me, and I strive for mathematical beauty rather than solving problems for the so called common good and evolution of humanity.

@mindhunter

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