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Find constants $a$ and $b$ such that $$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$$

First,$a$ should be positive to make sure the limit is meaningful as $x \to 0^-$ .

Then I check the limit of the numerator,say $- \frac{a}{2}<x<0$. For t in the interval(x,0), there's $\frac{1}{\sqrt{a+t}}<M$, where $M>0$,so $$| \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} | <M |\int^x_0t^2dt| =M \frac{-x^3}{3},$$ by using the sandwich theorem I get $\lim_{x \to 0^-} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} =0$,thing should be the same when $x>0$ .

So the limit is the form $0/0$ . Apply L'Hopital's rule twice I get $$\begin{align} \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} &= \lim_{x \to 0} \frac{x^2}{ \sqrt{a+x}(bx-\sin x)} \\ &= \lim_{x \to 0} \frac{2x} { \frac{bx-\sin x}{2 \sqrt{a+x}}+ \sqrt{a+x} (b- \cos x)} \\ &= \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3bx+2ab-2(a+x)\cos x -\sin x} \end{align}$$Now the limit becomes $ \frac{0}{2ab-2a}$,so $2a(b-1)=0$ and $b=1$ cause $a$ is positive. Apply L'Hopital's rule again $$\begin{align} \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3x+2a-2(a+x)\cos x -\sin x} &= \lim_{x \to 0} \frac{4( \sqrt{a+x} + \frac{x}{2 \sqrt{a+x}})}{3+2a \sin x- 2 \cos x + 2 x \sin x - \cos x} \\ & = \lim_{x \to 0} \frac{2(2a+3x)}{\sqrt{a+x} (3-3 \cos x + 2a \sin x + 2x \sin x )} \\ &= \frac{4a}{\sqrt{a} 0} \quad ?!\end{align}$$

and I fail to solve it.

Thanks in advance.

Michael Hardy
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Detective King
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2 Answers2

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Using the L'Hôpital's rule we have

$$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=\lim_{x \to 0}\frac{x^2}{(b-\cos x) \sqrt{a+x}}=\lim_{x \to 0} \frac{1}{\frac{(b-\cos x)}{x^2} \sqrt{a+x}}=1\\\iff (b=1)\land\left(\frac{\sqrt a}{2}=1\right)\iff (b=1)\land(a=4)$$

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Let's substitute $t\mapsto tx$: $$ \begin{align} \lim_{x\to0}\frac1{bx-\sin(x)}\int_0^x\frac{t^2\,\mathrm{d}t}{\sqrt{a+t}} &=\lim_{x\to0}\frac{x^3}{bx-\sin(x)}\color{#00A000}{\int_0^1\frac{t^2\,\mathrm{d}t}{\sqrt{a+tx}}}\\ &=\color{#00A000}{\frac1{3\sqrt{a}}}\lim_{x\to0}\frac{x^3}{bx-\sin(x)} \end{align} $$ If $b\ne1$, the limit on right would be $0$. Thus, we need $b=1$.

In this answer, it is shown that $\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$, therefore, we want $$ \frac1{3\sqrt{a}}\cdot6=1\implies a=4 $$

Community
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robjohn
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  • I'd like the idea you solve the problem. But I haven't learned how to solve $\lim_{x \to 0} \int_0^1\frac{t^2,\mathrm{d}t}{\sqrt{a+xt}}= \frac{1}{3\sqrt{a}}$ could you give more detail about this? – Detective King Mar 17 '14 at 20:12
  • Oh I think I've figured it out. $\lim_{x \to 0} \int_0^1\frac{t^2,\mathrm{d}t}{\sqrt{a+xt}}= \lim_{x \to 0} \frac{1}{\sqrt{a+cx}} \int_0^1 t^2,\mathrm{d}t= \frac{1}{3\sqrt{a}}$ – Detective King Mar 17 '14 at 20:21
  • @DetectiveKing: sort of... The function $\frac{t^2}{\sqrt{a+tx}}$ converges uniformly, monotonically, etc. to $\frac{t^2}{\sqrt{a}}$ as $x\to0$ so $\int_0^1\frac{t^2,\mathrm{d}t}{\sqrt{a+tx}}$ converges to $\int_0^1\frac{t^2,\mathrm{d}t}{\sqrt{a}}=\frac1{3\sqrt{a}}$. – robjohn Mar 17 '14 at 20:37