I need to calculate the limit without using L'Hôpital's rule:
limx→0arcsin(2x)ln(e−2x)−1
I know that: lima→0arcsinaa=1
But, how to apply this formula?
Answer
We have, limx→0arcsin2xln(e−2x)−1=limx→0arcsin2x2x2xln(e−2x)−1=limx→0arcsin2x2x−2xeln(1+(−2xe))×(−e)
This can be easily simplified to get the answer as −e. Hope it helps.
No comments:
Post a Comment