Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i'm learning by myself limits since my teachers doesn't respond any email and they are not making lectures, just pdf's... I'm trying to do my best, help pls). Also is there any good book or suggestion to learn limits .
\begin{align*}&\lim_{b \to a} \dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2} \cdot \dfrac{(a-\sqrt{a^2-b^2})}{(a-\sqrt{a^2-b^2})} \\=&
  \lim_{b \to a} \dfrac{a[a^2-(a^2-b^2)]-b^2(a-\sqrt{a^2-b^2})}{a[(a^2-\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
   = &\lim_{b \to a} \dfrac{-ab^2-ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
    =&\lim_{b \to a} \dfrac{-2ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
\end{align*}
 A: Notice: \begin{align}a \cdot (a + \sqrt{a^2-b^2}) - b^2 &= (a^2-b^2) +a\sqrt{a^2-b^2} \\&= 
\sqrt{a^2-b^2}(\sqrt{a^2-b^2} +a)\end{align}
and \begin{align}a \cdot (a - \sqrt{a^2-b^2}) - b^2 &= (a^2-b^2) -a\sqrt{a^2-b^2} \\ &= 
\sqrt{a^2-b^2}(\sqrt{a^2-b^2} -a)\end{align}
So  \begin{align} \lim_{b \to a} \dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}=  \lim_{b \to a} \dfrac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}-a}  \end{align}
Now it should be easy...
A: Let
$b = a-x$
where $x > 0$.
$\begin{array}\\
f(a, b)
&=\dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}\\
&=\dfrac{a \cdot (a + \sqrt{a^2-(a-x)^2}) - (a-x)^2}{a \cdot (a - \sqrt{a^2-(a-x)^2})-(a-x)^2}\\
&=\dfrac{a(a + \sqrt{2ax-x^2}) - a^2+2ax-x^2}{a(a - \sqrt{2ax-x^2})-a^2+2ax-x^2}\\
&=\dfrac{a\sqrt{2ax-x^2} +2ax-x^2}{- a\sqrt{2ax-x^2}+2ax-x^2}\\
&=\dfrac{a\sqrt{x}\sqrt{2a-x} +2ax-x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}\\
f(a, b)+1
&=\dfrac{a\sqrt{x}\sqrt{2a-x} +2ax-x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}+1\\
&=\dfrac{a\sqrt{x}\sqrt{2a-x} +2ax-x^2- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}\\
&=\dfrac{4ax-2x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}\\
&=\dfrac{4a\sqrt{x}-2x^{3/2}}{- a\sqrt{2a-x}+2a\sqrt{x}-x^{3/2}}\\
&\to 0
\qquad\text{as }x \to 0\\
\end{array}
$
A: Disclaimer : This method is probably overkill, you can make do with simpler arguments on limits.
What you want to do is find estimates of the numerator and denominator.
The main tool here is the following estimate for $y \rightarrow 0$ : $\sqrt{1+y} = 1 + O(y) $, where $O(y^r)$ means : something that is comparable to or smaller than $y^r$ as $y$ tends to $0$.
For simplicity, we will write $x= b-a$. Note that $x<0$.
So here, the numerator can be estimated as : $a^2 + a\sqrt{a^2-(a+x)^2}-(a+x)^2 = a^2 + a^2\sqrt{1-(1+\frac{x}a)^2}-(a+x)^2 $
$$\begin{aligned}&= a^2 + a^2\sqrt{2\frac{-x}{a} + O(x^2)} - a^2 + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x + O(x^2)}  + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x} \sqrt{1 + O(x)} + O(x)\\&= \sqrt{2}a^{\frac32} \sqrt{-x} (1+ O(x)) + O(x)\\&=  \sqrt{2}a^{\frac{3}32} \sqrt{-x} + O(x\sqrt{-x}) + O(x)\\&=  \sqrt{2}a^{\frac32} \sqrt{-x} + O(x)\end{aligned} $$
Similar computations tells us that the denominator is equivalent to $ -\sqrt{2}a^{\frac32} \sqrt{-x} $.
So the limit should be $-1$.
