How do you evaluate the following limit?
limx→0xsin−1xx−sinx
When I is L'Hopital's rule twice, I get:
limx→0(x2+2)cscx(1−x2)3/2
Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.
So, how do I find this limit?
Answer
limx→0xsin−1xx−sinx=limx→0sin−1xx⋅limx→01x−sinxx3⋅limx→01x
The first & the last limits are elementary and
for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion
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