Asked by Nuhu
Find limit x->x/2 tan^ sqroot (2sin^x+3sinx+4-sin^x+6sinx+2)
Answers
Answered by
Steve
You have very strange notation. I assume you mean
limit as x -> π/2 of
tan^2 √(2sin^2(x) + 3sinx + 4 - sin^2(x) + 6sinx + 2)
Clearly there is something wrong here. Try clearing it up, ok?
If I got it right, we have
tan^2(√(sin^2(x) - 3sinx + 2))
= tan^2(√((sinx-1)(sinx-2))))
as x -> π/2, that is
tan^2(0) = 0
So, I guess not.
limit as x -> π/2 of
tan^2 √(2sin^2(x) + 3sinx + 4 - sin^2(x) + 6sinx + 2)
Clearly there is something wrong here. Try clearing it up, ok?
If I got it right, we have
tan^2(√(sin^2(x) - 3sinx + 2))
= tan^2(√((sinx-1)(sinx-2))))
as x -> π/2, that is
tan^2(0) = 0
So, I guess not.
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