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?
-((a + Sqrt[a^2 - x])/(a (a + Sqrt[a^2 + x])))
, so L'Hôpital's rule is not strictly necessary. $\endgroup$