find the lim #1

$\begin{aligned} L & = \lim_{x \to 1} \frac {\ln(x^2+x-1}{x^2-1} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 1} \frac {\frac {2x+1}{x^2+x-1}}{2x} & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ & = \frac 32 = \boxed{1.5} \end{aligned}$

Let $x-1=y$ . Then the given expression equals $\dfrac {\ln (y^2+3y+1)}{y^2+2y}$ .

When $x\to 1, y\to 0$ , and the expression assumes the form $\dfrac {0}{0}$ .

So, applying L'Hospital's rule, we get the limit as $\dfrac {2\times 0+3}{(0^2+3\times 0+1)(2\times 0+2)}$

$=\dfrac {3}{2}=\boxed {1.5}$ .