limx→4(π6−arcsin(√x4)3√2x−7−1)
Hello! I need to solve this limit. I had solved it with the rule of L'Hôpital, but i can't without it. I tried multiplication by conjugate expression and using Special Limits. Please help me, I must solve it using only Special Limits and simple transformations. I can't use derivatives.
Answer
Well, let's put u=arcsin(√x/4) so that as x→4 we have u→π/6 and x=16sin2u. The expression under limit can be written as 116⋅π/6−u1/2−sinu⋅11/2+sinu⋅4−x3√2x−7−1
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