Question

$\lim_{n\to\infty} \frac{\int_{1/(n+1)}^{1/n} \tan^{-1}(nx)dx}{\int_{1/(n+1)}^{1/n} \sin^{-1}(nx)dx}$ is equal to

• A. 1
• B. 2
• C. 1/2
• D. 1/4

Answer 1

Answer: (C)

$\frac{1}{2}$

Answer 2

Let the given limit be $L$. $L = \lim_{n\to\infty} \frac{\int_{1/(n+1)}^{1/n} \tan^{-1}(nx)dx}{\int_{1/(n+1)}^{1/n} \sin^{-1}(nx)dx}$ We use the substitution $u = nx$, so $du = n \, dx$, which means $dx = \frac{du}{n}$. When $x = \frac{1}{n+1}$, $u = \frac{n}{n+1}$. When $x = \frac{1}{n}$, $u = 1$.

The integrals become: $\int_{1/(n+1)}^{1/n} \tan^{-1}(nx)dx = \int_{n/(n+1)}^{1} \tan^{-1}(u) \frac{du}{n} = \frac{1}{n} \int_{n/(n+1)}^{1} \tan^{-1}(u) du$ $\int_{1/(n+1)}^{1/n} \sin^{-1}(nx)dx = \int_{n/(n+1)}^{1} \sin^{-1}(u) \frac{du}{n} = \frac{1}{n} \int_{n/(n+1)}^{1} \sin^{-1}(u) du$

The limit expression simplifies to: $L = \lim_{n\to\infty} \frac{\frac{1}{n} \int_{n/(n+1)}^{1} \tan^{-1}(u) du}{\frac{1}{n} \int_{n/(n+1)}^{1} \sin^{-1}(u) du} = \lim_{n\to\infty} \frac{\int_{n/(n+1)}^{1} \tan^{-1}(u) du}{\int_{n/(n+1)}^{1} \sin^{-1}(u) du}$ Let $a_n = \frac{n}{n+1}$. As $n \to \infty$, $a_n \to 1$. The limit is of the form $\frac{0}{0}$. $L = \lim_{a_n \to 1^-} \frac{\int_{a_n}^{1} \tan^{-1}(u) du}{\int_{a_n}^{1} \sin^{-1}(u) du}$ Applying L'Hopital's Rule: $L = \lim_{a_n \to 1^-} \frac{\frac{d}{da_n}\int_{a_n}^{1} \tan^{-1}(u) du}{\frac{d}{da_n}\int_{a_n}^{1} \sin^{-1}(u) du} = \lim_{a_n \to 1^-} \frac{-\tan^{-1}(a_n)}{-\sin^{-1}(a_n)} = \lim_{a_n \to 1^-} \frac{\tan^{-1}(a_n)}{\sin^{-1}(a_n)}$ Substituting $a_n = 1$: $L = \frac{\tan^{-1}(1)}{\sin^{-1}(1)} = \frac{\pi/4}{\pi/2} = \frac{1}{2}$