Answer

Verified

486.6k+ 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.

Recently Updated Pages

Who among the following was the religious guru of class 7 social science CBSE

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Trending doubts

Derive an expression for drift velocity of free electrons class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE

The energy of a charged conductor is given by the expression class 12 physics CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Derive an expression for electric field intensity due class 12 physics CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Derive an expression for electric potential at point class 12 physics CBSE