Why can we substitute $x=0$ into $\frac{1-\sqrt{1-4x}}{2x}$?

Question

I am reading textbook A Walk Through Combinatorics. The section 8.1.2.1 is about Catalan numbers.
Let $c_n$ is Catalan numbers and $C(x)=\sum_{n\geq 0}c_nx^n$ is its generating function.
At the end of proof, the writer gives us the equation $$C(x)-1=xC(x)\cdot C(x)$$ then we have 2 solutions $$\frac{1+\sqrt{1-4x}}{2x},\frac{1-\sqrt{1-4x}}{2x}$$ I know that $C(x)$ has constant term $1$ because of $c_0=1$. (please correct me if I am wrong)
Then writer says "substituing $x=0$, we see thaht the second soluton has this property", but the denominator is $0$ if $x=0$. Why can we substitute $x=0$ into $\frac{1-\sqrt{1-4x}}{2x}$?

My understanding is to substitute $x\to 0$, then $$\lim_{x\to 0}\frac{1+\sqrt{1-4x}}{2x} \text{ does not exist.}$$ $$\lim_{x\to 0}\frac{1-\sqrt{1-4x}}{2x}=\frac{-\frac{-4}{2\sqrt{1-4x}}}{2}=1$$ Is it right?

Answer 1

The inference$$xf^2-f+1\implies f=\frac{1\pm\sqrt{1-4x}}{2x}$$assumes $x\ne0$ (otherwise we're not solving a quadratic), whereas the case $x=0$ clearly achieves $-f(0)+1=0$, which is correct. So the continuity of $f$ at $0$ implies in general$$f(x)=\lim_{y\to x}\frac{1-\sqrt{1-4y}}{2y}$$(once we've used the $x\to0$ limit to determine the $\pm$ sign should be $-$). As @AnginaSeng notes, we can evaluate this limit as $\frac{2}{1+\sqrt{1-4x}}$.

Answer 2

Yes you are right. Just a small remark: the first limit does not exist. You may see that if you take negative and positive values of $x$.

Answer 3

Using the "difference of two squares" trick $$\frac{1-\sqrt{1-4x}}{2x}=\frac{1-(1-4x)}{2x(1+\sqrt{1-4x})} =\frac{4x}{2x(1+\sqrt{1-4x})}=\frac2{1+\sqrt{1-4x}}.$$ There is no difficulty in substituting $x=0$ here.

Answer 4

This is a case for the rule of l'Hospital because it is

0/0

g(y)=1−Sqrt[1−4]

can be evaluated at y=0 and gives 0.

So all that is to do is develop the square root for small y.

 g(y) approx. 2 y - 2 y^2 # O(y^3) for small y

The rule of l'Hospital allows use to calculate with the approximate representation:

(2y+2y^2)/2y = 1+y for small y

So the

=1

exists and is equal 1.

The other series for small y is:

  1 + Sqrt[1 - 4 y] approx.  2-2y-2y^2+O(y^3) for small y

So the limit of the nominator is 2 and l'Hospital does not apply. But we can look more easily in the vicinity of zero for the function and see that the function diverges towards +∞. And we can infere that the divergence is of order y-1.

Since the equation chain of @angina-seng is correct, but the consequnce is that we have 1 again since we appointed to take the positive root in this equation chain only.

These are Taylor series expansions;