lim _(x → 1) ( (∫₀^(( x - 1 )²) t cos ( t² )dt)/(( x - 1 )sin ( x - 1 )) )

Solution

Maths sir · 4 steps

Put h = x - 1 and do the integral exactly: the top shrinks like h⁴, the bottom only like h².

1. Shift the variable. Let h = x - 1, so h → 0. The limit becomes (∫₀^(h²) t cos(t²) dt)/(h sin h), which is of the form 0/0.
2. Evaluate the integral. With u = t², du = 2t dt, so ∫ t cos(t²) dt = 1/2 sin(t²). Between 0 and h² this gives 1/2 sin(h⁴).

   (1/2 sin(h⁴))/(h sin h) = 1/2·sin(h⁴)/h⁴·h/(sin h)· h²

3. Take the limit. As h → 0, sin(h⁴)/h⁴ → 1 and h/(sin h) → 1. The extra factor h² → 0. This holds from both sides, because h sin h > 0 for small h ≠ 0.
4. So. The limit is 1/2 × 1 × 1 × 0 = 0, which is option (a).

The idea to keep. In a 0/0 limit, find the power of h that the top and the bottom each shrink like. If the top has the higher power, the limit is 0.

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