Find limit x->x/2 tan^ sqroot (2sin^x+3sinx+4-sin^x+6sinx+2)

Answer 1

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.