Without using integration and a graphing calculator, plot the graph of $y=f(x)$, given that its derivative is $f'(x)=e^{-x^{2}}$ and $f(0)=0$. With my little understanding of calculus, I calculated $\lim\limits_{x\to+\infty} f'(x)=0$ and $\lim\limits_{x\to-\infty} f'(x)=0$. Based on this information, I guessed that the graph must flatten for extremely large(whether positive or negative) values of $x$. Further, $$f''(x)=-2xe^{-x^{2}}$$ From this, I deduced that for $x<0$ the slope is increasing, while for $x>0$, the slope is decreasing. Given that $f(0)=0$ and $f'(0)=1$, the graph passes through the origin. Based on all this information, I figured the graph looks like this 
The only part I failed to figure out is this:

How to calculate the value of horizontal asymptotes enclosing the graph? Can this be done without explicitly involving integration?

 A: DISCLAIMER: This is done by integration, so count it as "additional information" as said in the comments.
$$f'(x)=e^{-x^2}$$
$$\int_0^{f(x)} \,d(f(x)) = \int_0^x e^{-x^2}\,dx $$
$$f(x)=\int_0^x e^{-x^2}\,dx$$
The function $y=e^{-x^2}$ should be familiar to you. It is the curve that defines the normal or gaussian distribution function.
$y=e^{-x^2}$">
Let's first look at the integral from $x=-\infty$ to $\infty$, rather than $x=0$ to $x$
In $\mathbb{R}^3$ consider two curves: $z=e^{-x^2}$ and $z=e^{-y^2}$.
Both of these stand vertically upon the x-y plane, with their peaks pointing in the direction of the z-axis.

The two areas under these curves would be the same: $A=\int_{-\infty}^{\infty} e^{-x^2}\,dx$ and $A=\int_{-\infty}^{\infty} e^{-y^2}\,dy$
Multiplying the two will give the volume of a 3-Dimensional bell-shaped curve as shown above (the right hand sketch)
We get :
$$A^2=\int_{-\infty}^{\infty} e^{-x^2}\,dx \times \int_{-\infty}^{\infty} e^{-y^2}\,dy$$
This gives a nested double integral:
$$A^2= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy$$ 
Now we can shift to polar co-ordinates ($r,\theta$), where r is polar radius vector of any point (x,y)
Let us view that 3-D bell curve as a solid of revolution created by rotating the curve $z=e^{-r^2}$, stretching from $r=0$ to $r=\infty$ about the z-axis, through an angle equal to $2\pi$ radians. We will have to change the limits of our double integrals accordingly.
We will also have to redefine that double integral from  $$  &  $$  to  $$  &  $$ , to adapt to a polar world. In the Cartesian World, the volume of a 3D solid is computed by adding the volumes of an infinite number of thin, vertical columns of square cross-section  $\,\,$. In the polar world, the same solid has to be imagined as being made of an infinite number of very tiny, concentric arc-shaped sections of radial thickness  $$  and arc-width  $$

Our new equation will then be:-
$$A^2= \int_0^{2\pi} \int_{0}^{\infty} re^{-r^2} \,d{\theta}\,dr$$
$$A^2=\left( \int_0^{2\pi} \,d{\theta} \right) \left( \int_0^\infty re^{-r^2}\,dr \right)$$
$$A^2=(2\pi) \left( \int_0^{\infty} \frac{1}{2} e^{-u}\,du \right) $$
$$A^2=\pi$$
$$A=\sqrt\pi$$
So, $$\int_0^\infty e^{-x^2}\,dx=\sqrt\pi$$
Now, to calculate the integral we require, let's resume at equation $A^2=\int_0^{2\pi} \int_0^\infty re^{-r^2} \,d\theta \,dr$
Let us switch limits of $r$ to : $r=0$ to $r=x$
We get the volume of our 3D solid to be :
$$A^2=\left( \int_0^{2\pi} \,d{\theta} \right) \left( \int_0^x re^{-r^2}\,dr \right)$$
$$=(2\pi)\left( \frac{1}{2} \int_0^{x^2} e^{-u}\,du \right)$$
$$A^2=\pi \times (1-e^{-x^2})$$
$$A=\sqrt{\pi (1-e^{-x^2})}$$
Now notice that the volume used here was the volume of our 3D bell shaped solid, but truncated at a radius of $x$. Something like:

So, the $A$ is actually $\int_{-x}^x e^{-x^2}\,dx$
But by the circular symmetry, $\int_0^x e^{-x^2}\,dx$ will be half of this $A$
Thus, finally:
$$\int_0^{x} e^{-x^2} \,dx = \frac{1}{2} \sqrt{\pi (1-e^{-x^2})} = f(x) $$
NOTE: this is only for $x>0$. For $x<0$ consider the negative part while taking the square root of $A^2$.
So the asymptotes can be found by applying the $\lim_{x \to \pm\infty}$ on $f(x)$
NOTE: for $x<0$, $f(x)=- \frac{1}{2} \sqrt{\pi (1-e^{-x^2})}$
A: 
How to calculate the value of horizontal asymptotes enclosing the graph? Can this be done without explicitly involving integration?

In other words: evaluate $\int_0^\infty \exp(-x^2)\;dx$ without integration. I assume "without integration" allows neither multiple integration nor contour integration and residue theory. The only way to do this that I can think of is: look it up.

Relating to the contour integration method:
Desbrow, Darrell.
On Evaluating $\displaystyle\int_{-\infty}^\infty e^{ax(x-2b)}\, dx$ by Contour Integration Round a Parallelogram. 
Amer. Math. Monthly $105\, (1998)$, no. $8,\, 726–731$.
