Evaluate limit

Question

Help me please ASAP thank you very much sirs

Answer 1

1. Do a reverse "difference of two squares" by multiplying above and below by by (√(3x + 1) + 1). The x will cancel and the limit comes out at 3/2.

2. Limit [x -> 0] f(x), where f(x) = sin{(π + tan(x))/(tan(x) - 2sec(x))}

As x -> 0, π + tan(x) -> π + 0 = π

tan(x) - 2sec(x) = tan(x) – 2/cos(x)

As x ->  0, tan(x) – 2/cos(x) -> 0 – 2/1 = -2

As x =-> 0, sin{(π + tan(x))/(tan(x) - 2sec(x))} -> sin(π/(-2)) = -sin(π/2) = -1

Answer: Limit = -1