This quiz is the LIMITing factor of our knowledge

Let $u=x^2-a^2$ . Then by Maclaurin series :

$\begin{aligned} \lim_{u \to 0} \frac u{1+u} & = \lim_{u \to 0} \frac u{u - \frac {u^2}2+ \frac {u^3}3 - \cdots } & \small \blue{\text{for }|u| < 1} \\ & = \lim_{u \to 0} \frac 1{1 - \frac u2+ \frac {u^2}3 - \cdots } & \small \blue{\text{Divide up and down by }u} \\ & = \boxed 1 \end{aligned}$

Let $y = x^2-a^2$ . Then as $x\to a$ , $y\to 0$ . The limit becomes $\lim_{y\to0} \frac y{\ln(y+1)} .$ Apply L'Hôpital's rule gives $\lim_{y\to0} \frac1{1/(y+1)} = \lim_{y\to0} (y+1) = \boxed1.$

We can use L'Hopital's Rule here because when we plug $a$ in we get a $\Large \frac{0}{0}$ situation: $\begin{aligned} \lim\limits_{x\to a}\frac{x^2-a^2}{\ln(x^2-a^2+1)} \Longrightarrow \lim\limits_{x\to a}\frac{\frac{d}{dx}(x^2-a^2)}{\frac{d}{dx}(\ln(x^2-a^2+1))} &= \lim\limits_{x\to a}\frac{\sout{2x}}{\frac{\sout{2x}}{x^2-a^2+1}} \\ &= \lim\limits_{x\to a} x^2-a^2+1 \\ &= a^2 - a^2 + 1 = \boxed{1} \end{aligned}$

We have the standard limit :

$\displaystyle \lim_{h\to 0} \dfrac {h}{\ln (1+h)}=\displaystyle \lim_{h\to 0} \dfrac {\ln(1+h)}{h}=1$

Letting $x^2-a^2=h$ , we get the required limit as $\boxed 1$ .

"Fakesolve" solution:

Let $a = 0$ . The limit simplifies to $\displaystyle \lim_{x\rightarrow 0} \frac{x^2}{\ln(x^2+1)}$ . Directly substituting $x=0$ results in an indeterminate form where we can use L'Hopital's Rule. Then:

$\newcommand\myeq{\stackrel{\mathclap{\mbox{L. H.}}}{=}} \begin{aligned} \displaystyle \lim_{x\rightarrow 0} \frac{x^2}{\ln(x^2+1)} \text{ } &\myeq \text{ } \lim_{x\rightarrow 0} \displaystyle \frac{2x}{\left(\frac{2x}{x^2+1}\right)}\\ &= \displaystyle \lim_{x\rightarrow 0} x^2 + 1\\ &= \boxed{1} \end{aligned}$