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Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 98

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Find $\displaystyle \lim_{t \to 0^+} \frac{A(t)}{B(t)}$

$\displaystyle A(t) = \int^t_0 \sin (x^2) dx$ while $B(t)$, using the area of triangle, $\displaystyle B(t) = \frac{1}{2} (t) \left( \sin (t^2) \right)$

So,

$\displaystyle \lim_{t \to 0^+} \frac{A(t)}{B(t)} = \lim_{t \to 0} \frac{\int^t_0 \sin (x^2) dx}{\frac{1}{2}t \sin (t^2)} = \lim_{t \to 0} \frac{2 \int^t_0 \sin(x^2)dx}{t \sin (t^2)}$

By applying L'Hospital's Rule...

$ \begin{equation} \begin{aligned} \lim_{t \to 0} \frac{2 \int^t_0 \sin(x^2)dx}{t \sin (t^2)}&= \lim_{t \to 0} \frac{2 \frac{d}{dt}\left( \int^t_0 \sin (x^2) dx \right)}{\frac{d}{dt}\left( t \sin (t^2) \right)}\\ \\ &= \lim_{t \to 0} \frac{2 \sin (t^2)}{t\left( \cos (t^2) \right)(2t)+(1)\sin (t^2)}\\ \\ &= \lim_{t \to 0} \frac{2 \sin (t^2) }{ 2t^2 \cos t^2 + \sin t^2} \end{aligned} \end{equation} $

If we evaluate the limit, we will still get an indeterminate form, so we must apply L'Hospital's Rule once more. Thus,

$ \begin{equation} \begin{aligned} \lim_{t \to 0} \frac{2 \sin (t^2) }{ 2t^2 \cos t^2 + \sin t^2} &= \lim_{t \to 0} \frac{2 \cos (t^2)(2t)}{2 \left[ t^2 (-\sin t^2) (2t) + (2t)\cos t^2 \right] + (\cos t^2)(2t)}\\ \\ &= \lim_{t \to 0} \frac{4t \cos t^2}{-4 t^3 \sin t^2 + 4t \cos t^2 + 2t \cos t^2}\\ \\ &= \lim_{t \to 0} \frac{2t\left(2\cos t^2\right)}{2t \left( -2t^2 \sin t^2 + 2 \cos t^2 + \cos t^2\right)}\\ \\ &= \lim_{t \to 0} \frac{2 \cos t^2}{3 \cos t^2 - 2 t^2 \sin t^2}\\ \\ &= \frac{2\cos (0)^2}{3 \cos (0)^2 - 2(0)^2 \sin (0)^2}\\ \\ &= \frac{2(1)}{3(1) - 0}\\ \\ &= \frac{2}{3} \end{aligned} \end{equation} $