# Problem with limit that requires L'Hôpital's rule to compute

Consider the following limit.

Limit[(a - Sqrt[a^2 + x])/(a^2 - a*Sqrt[a^2 - x]), x -> 0, Assumptions -> {a > 0}]

Mathematica 9.0.1.0 gives -1/a, which is the correct answer. Notice that this limit is not trivial to compute, because both the numerator and denominator vanish when x=0. Therefore, L'Hopital's rule is required here.

Now remove the assumption.

Limit[(a - Sqrt[a^2 + x])/(a^2 - a*Sqrt[a^2 - x]), x -> 0]

For this, Mathematica gives 1/a, which is incorrect for general a (although it is correct for a<0).

Is this last result a bug, or am I missing something?

• By elementary algebra, the function is equivalent to -((a + Sqrt[a^2 - x])/(a (a + Sqrt[a^2 + x]))), so L'Hôpital's rule is not strictly necessary. Commented Feb 9, 2014 at 19:00

When Assumptions -> {a > 0} is used, you get the correct limit. But when no assumptions are placed, Mathematica tries to evaluate the limit for a general complex $a$. This second result is not correct for $\Re(a) > 0$: The correct limit is $-1/a$ for $\Re(a) > 0$, and $1/a$ for $\Re(a) < 0$.