lim _(x → 0) (∫₀ˣ t sin ( 10t )dt)/x is equal to

Solution

Maths sir · 4 steps

Top and bottom both go to 0, so differentiate both, and the integral sign falls away.

1. Spot the form. At x = 0 the integral runs from 0 to 0, so the numerator is 0. The denominator x is also 0, so this is a 0/0 form and L'Hospital's rule applies.
2. Differentiate both. By the fundamental theorem of calculus, d/dx∫₀ˣ t sin(10t) dt = x sin(10x). The derivative of the denominator x is 1.

   lim(x → 0)(∫₀ˣ t sin(10t) dt)/x = lim(x → 0)(x sin(10x))/1 = 0 × sin 0 = 0

3. Quick check. Near 0, sin(10t) ≈ 10t, so the integrand is about 10t² and the integral is about 10x³/3. Divided by x this is 10x²/3, which goes to 0.
4. So. The limit is 0, which is option (c).

The idea to keep. When an integral with a variable upper limit sits in a 0/0 limit, use L'Hospital: the derivative of ∫₀ˣ f(t) dt is just f(x). No need to integrate at all.

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