The value of lim _(h → 0) 2{ (√3 sin ( π/6 + h ) - cos ( π/6 + h ))/(√3 h( √3 cosh - sinh )) } is :

Solution

Maths sir · 4 steps

Fold the numerator into one sine: it collapses to 2 sin h, and then the standard limit finishes it.

1. Read the notation. Here cosh and sinh mean cos h and sin h, not hyperbolic functions. As h → 0, the numerator and the denominator both go to 0, so we need to simplify first.
2. Combine the numerator. Take 2 common and use √3/2 = cos π/6 and 1/2 = sin π/6. With x = π/6 + h, the numerator is 2(sin x cos π/6 - cos x sin π/6) = 2 sin(x - π/6) = 2 sin h.

   lim(h → 0) 2 · (2 sin h)/(√3 h(√3 cos h - sin h)) = 4/√3 · lim(h → 0)(sin h)/h · lim(h → 0) 1/(√3 cos h - sin h)

3. Take the limits. (sin h)/h → 1, and the bracket √3 cos h - sin h → √3 × 1 - 0 = √3. So the limit is 4/√3 × 1 × 1/√3 = 4/3.
4. So. The value is 4/3, option (a).

The idea to keep. Whenever you see a sin x ± b cos x, take √(a²+b²) common and write it as a single sine. A messy 0/0 usually turns into (sin h)/h.

Answer from the NTA final answer key. Solution by PrepWiser.