# 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?

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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. – Michael E2 Feb 9 '14 at 19:00

Manipulate[Plot[(a - Sqrt[a^2 + x])/(a^2 - a*Sqrt[a^2 - x]), {x, -1, 1}], {a, -3, 3}]

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 $\Re(a) < 0$.