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Find constants so the limit of an integral has a prescribed value

by RoRi

Find values for the constants a and b so that

    \[ \lim_{x \to 0} \frac{1}{bx - \sin x} \int_0^x \frac{t^2 \, dt}{\sqrt{a+t}} = 1. \]

First, since the integral \int_0^x \frac{t^2 \, dt}{\sqrt{a+t}} is continuous (Theorem 3.4 of Apostol) we know

    \[ \lim_{x \to 0} \int_0^x \frac{t^2 \, dt}{\sqrt{a+t}} = \int_0^0 \frac{t^2 \, dt}{\sqrt{a+t}} = 0. \]

We also have

    \[ \lim_{x \to 0} (bx - \sin x) = 0. \]

Hence, we may apply L’Hopital’s rule to the quotient giving us (using the First Fundamental Theorem of Calculus for the numerator)

    \[ 1 = \lim_{x \to 0} \frac{\int_0^x \frac{t^2 \, d}{\sqrt{a+t}}}{bx - \sin x} = \lim_{x \to 0} \frac{ \frac{x^2}{\sqrt{a+x}}}{b - \cos x}. \]

Since the limit as x \to 0 of the numerator is 0, we must have the limit of the denominator equal to 0 as well (otherwise the whole limit would be 0 instead of 1).

    \[ \lim_{x \to 0} (b - \cos x) = 0 \quad \implies \quad b - \cos 0 = 0 \quad \implies \quad b = 1. \]

Now, substituting b = 1 back into our limit and using L’Hopital’s two more times (since again we have the indeterminate form 0/0),

    \begin{align*} \lim_{x \to 0} \frac{\frac{x^2}{\sqrt{a+x}}}{1 - \cos x} &= \lim_{x \to 0} \frac{2x}{(\sin x)\sqrt{a+x} + \frac{1 - \cos x}{2 \sqrt{a+x}}} \\[9pt] &= \lim_{x \to 0} \frac{2}{\cos x \sqrt{a+x} + \frac{\sin x}{2 \sqrt{x+a}} + \frac{\sin x}{2 \sqrt{x+a}} - \frac{1 - \cos x}{4 \sqrt{x+a}}} \\[9pt] &= \frac{2}{\sqrt{a}}. \end{align*}

Therefore,

    \[ \frac{2}{\sqrt{a}} = 1 \quad \implies \quad a = 4. \]

Hence, the requested constants are

    \[ a = 4, \qquad b = 1. \]

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