Question

Which of the following is/are true:
Statement-1: \underset{x\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}\left\\{ x \right\\} does not exist.
Statement-2: \left\\{ x \right\\} is discontinuous at $x=0$ (where \left\\{ . \right\\} denotes fractional part function).
(a) Both statement 1 and 2 are true and statement 2 is the correct explanation of statement 1.
(b) Both statement 1 and 2 are true but statement 2 is not the correct explanation of statement 1.
(c) Statement 1 true but statement 2 is false.
(d) Statement 1 false but statement 2 is true.

Answer 1

Hint: Check the validity of the two given statements by evaluating the limits using the properties of fractional part function and sine function.

Complete step-by-step answer:

We will first check the limit for the function \left\\{ x \right\\} around the point $x=0$.
We have the function \left\\{ x \right\\}. This function is the fractional part function. It returns the value of the fractional part of a real number. Its value lies between $0$ and $1$ as it gives only fractional values.

We know that $\\{x\\}<0$ for $x<0$ and $\\{x\\}>0$ for $x>0$.
We have $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\\{x\\}=-1$ and $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\\{x\\}=0$.

Thus we have \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left\\{ x \right\\}\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left\\{ x \right\\} as left side of the equation will have value $-1$ while right side of the equation will have value $0$.

Hence, the function \left\\{ x \right\\} is discontinuous around the point $x=0$.

Thus, statement 2 is true.

Now, we will check the validity of the first statement.

As we have $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\\{x\\}=-1$ and $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\\{x\\}=0$ thus, we have $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{\sin }^{-1}}\\{x\\}={{\sin }^{-1}}\left( -1 \right)=\dfrac{-\pi }{2}$ and $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{\sin }^{-1}}\\{x\\}={{\sin }^{-1}}\left( 0 \right)=0$.
So, we have \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{\sin }^{-1}}\left\\{ x \right\\}\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{\sin }^{-1}}\left\\{ x \right\\}
Hence, the limit of the function \underset{x\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}\left\\{ x \right\\} does not exist.

Thus, statement 1 is correct as well.

Also, statement 2 is the correct explanation of the statement 1.

Option (a) is the correct answer.

Note: Limit of a function is a fundamental concept in calculus that analyses the behaviour of that function around a point. The notation of a limit has many applications in modern calculus. We can use the limit of a function to check differentiability of the function around any point. We can also use the limit to integrate a function defined over an interval.
When we say that a function has limit L, it means that the function gets closer and closer to the value L around the point at which the limit is applied to.