# $\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]$ , where $\left[ . \right]$ denotes greatest integer function, is equal to

(a) 0

(b) 1

(c) Does not exist

(d) \[\sin 1\]

Answer

Verified

363.6k+ views

Hint: To evaluate the given limit, simplify the function using the formulas related to the greatest integer function and then apply the limits to the given function to get the value of the limit.

We have been provided the function $\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]$

We know that the function \[\left[ {} \right]\] stands for greatest integer function. This function rounds down a real number to the nearest integer. This function gives the value of the greatest integer that is less than or equal to the given real number.

We will evaluate the right hand and left hand limit of the given function.

Firstly, we will evaluate the left hand limit of the function.

We know that \[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\left[ x \right]=a\] and \[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\left[ x \right]=a-1\].

Thus, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}\].

As we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$. We get this limit with the use of L’Hopital Rule.

Thus, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}=1\].

Hence, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 1 \right]=1\].

Now, we will evaluate the left hand side of the limit.

Thus, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}\].

As we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$. We get this limit with the use of L’Hopital Rule.

Thus, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}=1\].

Hence, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ 1 \right]=1\].

Thus, we get the value of limit as \[\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=1\].

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.

Answer is Option (b)

Note: To evaluate the value of the limit of the function, it’s necessary to know about the greatest integer function. Simplify the terms involving greatest integer function using some of its properties to evaluate the limit of the function.

We have been provided the function $\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]$

We know that the function \[\left[ {} \right]\] stands for greatest integer function. This function rounds down a real number to the nearest integer. This function gives the value of the greatest integer that is less than or equal to the given real number.

We will evaluate the right hand and left hand limit of the given function.

Firstly, we will evaluate the left hand limit of the function.

We know that \[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\left[ x \right]=a\] and \[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\left[ x \right]=a-1\].

Thus, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}\].

As we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$. We get this limit with the use of L’Hopital Rule.

Thus, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}=1\].

Hence, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 1 \right]=1\].

Now, we will evaluate the left hand side of the limit.

Thus, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}\].

As we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$. We get this limit with the use of L’Hopital Rule.

Thus, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left[ x-3 \right]}{x-3}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\sin \left( x-3 \right)}{x-3}=1\].

Hence, we have \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ 1 \right]=1\].

Thus, we get the value of limit as \[\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\sin \left[ x-3 \right]}{x-3} \right]=1\].

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.

Answer is Option (b)

Note: To evaluate the value of the limit of the function, it’s necessary to know about the greatest integer function. Simplify the terms involving greatest integer function using some of its properties to evaluate the limit of the function.

Last updated date: 25th Sep 2023

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