I've been looking at Fermat's method of obtaining a tangent line through this resource https://cedar.wwu.edu/cgi/viewcontent.cgi?article=1012&context=wwu_honors and it says to basically take an $x_1$ after the $x$ you're interested in finding the tangent for so that $x_1=x+e$. Let $T(x)$ be the equation for the tangent, and let $(v,0)$ be the point where the tangent line intercepts the $x$-axis. So
$$\frac{f(x)}{x-v}=\frac{T(x_1)}{x+e-v},$$
and then you can say that
$$\frac{f(x)}{x-v}$$
is similar to
$$\frac{f(x+e)}{x+e-v},$$
and using $f(x)=x^2$ and $x=1$, trying to solve for $v$ you get
$$e +e^2= 2ev +ve^2$$
and dividing each side by $e$ you get
$$1+e=2v+ev.$$
Then setting $e =0: v=1/2$. so my question is why does this actually work, what in these operations is making it work because if you plug $e=0$ in the initial equation
$$\frac{f(x)}{x-v}$$
is similar to
$$\frac{f(x+e)}{x+e-v}$$
you won't get anywhere.