# Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 96

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Evaluate $\displaystyle \lim_{x \to 0} \frac{1}{x^3}\int^x_0 \sin (t^2) dt$

$\displaystyle \lim_{x \to 0} \frac{1}{x^3}\int^x_0 \sin (t^2) dt = \lim_{x \to 0} \frac{\int^x_0 \sin (t^2) dt}{x^3}$

By applying L'Hospital's Rule

$\displaystyle \lim_{x \to 0} \frac{\int^x_0 \sin (t^2) dt}{x^3} = \lim_{x \to 0} \frac{\frac{d}{dx} \left(\int^x_0 \sin(t^2) dt \right) }{\frac{d}{dx} (x^3)} = \lim_{x \to 0} \frac{\sin x^2}{3x^2}$

If we evaluate the limit, we will still get a indeterminate form, so we must apply the L'Hospital's Rule until the limit can be determined so...

\begin{aligned} \lim_{x \to 0} \frac{\sin x^2}{3x^2} = \lim_{x \to 0} \frac{2x\cos x^2}{6x} &= \lim_{x \to 0} \frac{2\cos x^2}{6}\\ \\ &= \frac{2\cos(0)^2}{6}\\ \\ &= \frac{2(1)}{6}\\ \\ &= \frac{1}{3} \end{aligned}