Question

Let $\lim_{x\to\pi}[\sin(\sin^{-1}x)]=\ell_1$ and $\lim_{x\to\pi/2}[\sin^{-1}(\sin x)]=\ell_2$, where

[.]=G.I.F. Then

• A. $\ell_1=1$, $\ell_2$ does not exist
• B. $\ell_1=0$, $\ell_2=1$
• C. $\ell_1$ does not exist, $\ell_2=1$
• D. Both $\ell_1$ and $\ell_2$ do not exist

Answer 1

Answer: (C)

C

Answer 2

To evaluate the limits $\ell_1$ and $\ell_2$, we need to analyze the domain of the functions and the behavior of the greatest integer function (G.I.F.).

Evaluation of $\ell_1 = \lim_{x\to\pi}[\sin(\sin^{-1}x)]$

1. Domain of the function: The function is $f(x) = \sin(\sin^{-1}x)$. The domain of $\sin^{-1}x$ is $[-1, 1]$. Therefore, the domain of $\sin(\sin^{-1}x)$ is also $[-1, 1]$.

2. Limit point: The limit is taken as $x \to \pi$. Since $\pi \approx 3.14159$, the value $\pi$ is outside the domain $[-1, 1]$ of the function $f(x)$.

3. Existence of limit: For a limit $\lim_{x\to a} g(x)$ to exist, the function $g(x)$ must be defined in a punctured neighborhood of $a$. In this case, as $x \to \pi$, $x$ is not in the domain of $\sin(\sin^{-1}x)$. Thus, the function is not defined in any neighborhood of $\pi$. Therefore, $\ell_1$ does not exist.

Evaluation of $\ell_2 = \lim_{x\to\pi/2}[\sin^{-1}(\sin x)]$

1. Understanding $\sin^{-1}(\sin x)$: The function $g(x) = \sin^{-1}(\sin x)$ is a periodic function with period $2\pi$. Its graph is a triangular wave.

   • For $x \in [-\pi/2, \pi/2]$, $\sin^{-1}(\sin x) = x$.
   • For $x \in [\pi/2, 3\pi/2]$, $\sin^{-1}(\sin x) = \pi - x$.

2. Evaluating the left-hand limit (LHL): As $x \to (\pi/2)^-$, $x$ is slightly less than $\pi/2$. In this interval, $\sin^{-1}(\sin x) = x$. So, $\lim_{x\to(\pi/2)^-} [\sin^{-1}(\sin x)] = \lim_{x\to(\pi/2)^-} [x]$. Since $\pi/2 \approx 1.5708$, as $x$ approaches $\pi/2$ from the left, $x$ is a value slightly less than $1.5708$ (e.g., $1.5707$). The greatest integer less than or equal to such a value is $1$. Therefore, $\lim_{x\to(\pi/2)^-} [x] = 1$.

3. Evaluating the right-hand limit (RHL): As $x \to (\pi/2)^+$, $x$ is slightly greater than $\pi/2$. In this interval, $\sin^{-1}(\sin x) = \pi - x$. So, $\lim_{x\to(\pi/2)^+} [\sin^{-1}(\sin x)] = \lim_{x\to(\pi/2)^+} [\pi - x]$. Let $x = \pi/2 + h$, where $h \to 0^+$. Then $\pi - x = \pi - (\pi/2 + h) = \pi/2 - h$. So, we need to evaluate $\lim_{h\to 0^+} [\pi/2 - h]$. As $h$ approaches $0$ from the positive side, $\pi/2 - h$ is a value slightly less than $\pi/2 \approx 1.5708$ (e.g., $1.5707$). The greatest integer less than or equal to such a value is $1$. Therefore, $\lim_{x\to(\pi/2)^+} [\pi - x] = 1$.

4. Conclusion for $\ell_2$: Since the LHL and RHL are equal ($\text{LHL} = 1$, $\text{RHL} = 1$), the limit $\ell_2$ exists and $\ell_2 = 1$.

Summary: $\ell_1$ does not exist. $\ell_2 = 1$.

This matches option C.