<s>[INST] 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*}
 [/INST] 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}
$
</s>