Q.$\displaystyle\lim_{x\to \pi/3}\left(\frac{\tan^2x-3}{\sec^3x-8}\right)=$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{4}$

Because sine and cosine are continuous everywhere, direct substitution always gives the limit of sin x or cos x at any point a. The single deepest result in this family is that the limit, as an angle theta measured in radians tends to zero, of sin(theta) divided by theta, equals exactly 1 — proved geometrically by squeezing the area of a small triangle, a circular sector, and a larger triangle between one another and letting the squeeze force the ratio to 1 from both the positive and negative side of zero. From this one theorem an entire family of corollaries follows immediately: the reciprocal ratio theta over sin(theta) also tends to 1, the corresponding tangent ratios tend to 1, and any of these ratios scaled by a constant p inside the argument still tends to 1. The Squeeze (or Sandwich) theorem underlying the proof states more generally that if one function is trapped between two others that both converge to the same limiting value, the trapped function must conve …