You can show geometrically that 1 < tan θ/θ < sec θ on (0, π/2); this is also true on (-π/2, 0) - not asked See the picture at https://math.stackexchange.com/questions/2586824/proving-theta-tan-theta-with-geometry where it is shown that sin θ < θ < tan θ - the area of OAK is sin θ/2, the area of the sector is θ/2, and the area of OAL is tan θ/2, and we see via the picture that the area of OAK is less than the area of the sector is less than the area of OAL. Multiplying by 2 proves the inequality. Then, if sin θ < θ < tan θ, clearly, tan θ/θ > 1, and

tan θ/θ = sin θ / cos θ / θ < θ/cos θ/θ = 1/cos θ = sec θ.

Then, trivially 1 has a limit of 1 as x -> 0, and sec θ has a limit of 1 as x -> 0 (recall that sec θ = 1/cos θ, and cos 0 = 1), so tan θ/θ has a limit of 1 as x -> 0.