Step 1: Spot the indeterminate form.
For $\lim_{x\to1}\dfrac{ab^x-a^xb}{x^2-1}$, putting $x=1$ gives $\dfrac{ab-ab}{0}=\dfrac00$. Rather than L'Hospital, let us substitute $x=1+h$ with $h\to0$ and use first-order expansions.
Step 2: Factor the denominator.
$x^2-1=(x-1)(x+1)$, so near $x=1$ the denominator is $h(2+h)$.
Step 3: Expand each exponential to first order.
Using $k^{1+h}=k\cdot k^{h}=k\,e^{h\ln k}\approx k(1+h\ln k)$, we get $b^{x}=b^{1+h}\approx b(1+h\ln b)$ and $a^{x}\approx a(1+h\ln a)$.
Step 4: Expand the numerator.
\[ ab^{x}-a^{x}b\approx ab(1+h\ln b)-ab(1+h\ln a)=ab\,h(\ln b-\ln a). \]
Step 5: Form the ratio and cancel $h$.
\[ \frac{ab\,h(\ln b-\ln a)}{h(2+h)}=\frac{ab(\ln b-\ln a)}{2+h}. \]
Step 6: Take the limit $h\to0$.
The expression tends to $\dfrac{ab}{2}(\ln b-\ln a)=\dfrac{ab}{2}\log\!\left(\dfrac{b}{a}\right)$, matching the key. \[ \boxed{\dfrac{ab}{2}\log\!\left(\dfrac{b}{a}\right)} \]