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.

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

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.