Verified answer

we know

as t->0 sin t/ t = 1

taking reciprocal t/sin t =1 (I used t for theta)

another way

it is of the form inf / inf

so using L'Hospitals rule

d(t)/dt//(d/dt(sin t)) = 1/cos t = 1

Your question were given problem in case you're trying to lim x to 0, something divided via 0 will be countless. enable's see for nominator merely: (a million- cosx)² = (a million - 2cosx + cos²x) = a million -2cosx + (a million - cos2x)/2 then, lim (a million- cosx)² = a million - 2 - 0 = -a million the position, cos2x = a million - 2con²x and cos0 = a million next, (cosxtanx) = cosx(sinx / cosx) = sinx then, lim (cosxtanx) = sin0 = 0

Apply L' Hospital rule

We get

lim theta->0 1/cos(theta)

Putting theta = 0, we get 1