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
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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.
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.
Last updated date: 24th Sep 2023
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