Total inductance of two parallel aiding inductors
Intuition tells us the answer is \$\small L\$, but it's always wise to keep an eye on intuition.

Thus, we have \$\frac{0}{0}\$, which is an indeterminate form.

One way out of the dilemma is L'hopital's rule:

$$ \lim_{k\to 1} \small \left (\frac{L^2-k^2L^2}{2L-2kL}\right )=\normalsize \lim_{k\to 1}\: \small \left (\frac{\large\frac{d}{dk}\small (L^2-k^2L^2)}{\large \frac{d}{dk}\small (2L-2kL)}\right)=\normalsize\lim_{k\to 1} \small \frac{-2k\:L^2}{-2L}=\normalsize L$$

No surprises; intuition checks out.