Question

If $I_1(\sqrt{2}) = \frac{9\pi}{16}$, then $\lim_{x\to 0}\frac{cos(I_1(x))}{x^2}$ is equal to -

• A. $-\frac{1}{4}$
• B. $-\frac{1}{8}$
• C. $\frac{1}{8}$
• D. 1

Answer 1

Answer: (B)

$-\frac{1}{8}$

Answer 2

1. Evaluate $I_1(x)$: $I_1(x) = \int\frac{x dx}{x^4+4}$ Let $u = x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. $I_1(x) = \int\frac{\frac{1}{2} du}{u^2+4} = \frac{1}{2} \int\frac{du}{u^2+2^2}$ Using the standard integral formula $\int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a}) + C$: $I_1(x) = \frac{1}{2} \cdot \frac{1}{2}\tan^{-1}(\frac{u}{2}) + C = \frac{1}{4}\tan^{-1}(\frac{x^2}{2}) + C$.

2. Determine the constant C: Given $I_1(\sqrt{2}) = \frac{9\pi}{16}$. Substitute $x=\sqrt{2}$ into $I_1(x)$: $I_1(\sqrt{2}) = \frac{1}{4}\tan^{-1}(\frac{(\sqrt{2})^2}{2}) + C = \frac{1}{4}\tan^{-1}(\frac{2}{2}) + C = \frac{1}{4}\tan^{-1}(1) + C$. Since $\tan^{-1}(1) = \frac{\pi}{4}$: $I_1(\sqrt{2}) = \frac{1}{4} \cdot \frac{\pi}{4} + C = \frac{\pi}{16} + C$. Equating this to the given value: $\frac{\pi}{16} + C = \frac{9\pi}{16} \implies C = \frac{8\pi}{16} = \frac{\pi}{2}$. So, $I_1(x) = \frac{1}{4}\tan^{-1}(\frac{x^2}{2}) + \frac{\pi}{2}$.

3. Evaluate the limit: We need to find $L = \lim_{x\to 0}\frac{\cos(I_1(x))}{x^2}$. As $x \to 0$, $I_1(x) \to \frac{1}{4}\tan^{-1}(0) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. So, as $x \to 0$, $\cos(I_1(x)) \to \cos(\frac{\pi}{2}) = 0$. The limit is of the indeterminate form $\frac{0}{0}$, so we can use L'Hopital's Rule. $L = \lim_{x\to 0}\frac{-\sin(I_1(x)) \cdot I_1'(x)}{2x}$. Recall that $I_1'(x)$ is the integrand of $I_1(x)$, so $I_1'(x) = \frac{x}{x^4+4}$. Substitute $I_1'(x)$: $L = \lim_{x\to 0}\frac{-\sin(I_1(x)) \cdot \frac{x}{x^4+4}}{2x}$. Simplify by cancelling $x$ (since $x \ne 0$ in the limit): $L = \lim_{x\to 0}\frac{-\sin(I_1(x))}{2(x^4+4)}$. Now, substitute $x=0$: As $x \to 0$, $I_1(x) \to \frac{\pi}{2}$, so $\sin(I_1(x)) \to \sin(\frac{\pi}{2}) = 1$. The denominator $2(x^4+4) \to 2(0^4+4) = 8$. Therefore, $L = \frac{-1}{8}$.

   Alternatively, using small angle approximation: Let $y = I_1(x)$. As $x \to 0$, $y \to \frac{\pi}{2}$. Let $y = \frac{\pi}{2} + \delta$, where $\delta = I_1(x) - \frac{\pi}{2} = \frac{1}{4}\tan^{-1}(\frac{x^2}{2})$. As $x \to 0$, $\frac{x^2}{2} \to 0$. For small $z$, $\tan^{-1}(z) \approx z$. So, $\delta \approx \frac{1}{4} \cdot \frac{x^2}{2} = \frac{x^2}{8}$. Now, $\cos(I_1(x)) = \cos(\frac{\pi}{2} + \delta) = -\sin(\delta)$. For small $\delta$, $\sin(\delta) \approx \delta$. So, $\cos(I_1(x)) \approx -\delta \approx -\frac{x^2}{8}$. The limit becomes $\lim_{x\to 0}\frac{-\frac{x^2}{8}}{x^2} = -\frac{1}{8}$.