Answer

Verified

452.4k+ views

Hint: The limit of a function exists only if left hand limit and right hand limit exist and both are equal.

Also, the value of the limit will be equal to the value of the right hand limit and (or) the left hand limit.

We know, the limit of a function exists only if the left hand limit and right hand limit exist and both are equal.

So, first, we will find the left hand limit of the function \[f(x)\] at \[x=1\].

We know, the left hand limit of a function\[f\left( x \right)\] at the point \[x=a\] is given as

\[\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)\]

So , the left hand limit of the function\[f\left( x \right)\] at the point\[x=1\] is given as

\[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)\]

Now, we know h is a very small quantity and subtracting h from 1 will give a number less than 1 but very close to 1. But, since the number is not equal to 1, so, \[f\left( 1-h \right)=1-h+2\]

So, \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,1-h+2=1-0+2=3\]

Now, we will find the right hand limit of the function\[f\left( x \right)\] at the point \[x=1\].

We know right hand limit of a function \[f\left( x \right)\] at \[x=a\] is given by

\[R.H.L=\underset{h\to 0}{\mathop{\lim }}\, f\left( a+h \right)\]

So, the right hand limit of the function \[f\left( x \right)\] at the point \[x=1\] is given by

\[\underset{h\to 0}{\mathop{\lim }}\, f\left( 1+h \right)\]

Now, we know h is a very small number. So, 1+h will be slightly greater than 1 and not equal to 1. So, \[f\left( 1+h \right)=1+h+2\]

So, \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+h+2 \right)=3\]

We can clearly see that both the left hand limit and the right hand limit of the function \[f\left( x \right)\] exist at the point \[x=1\].

Also, the left hand limit and the right hand limit of the function \[f\left( x \right)\] at the point \[x=1\] are equal.

Since, the value of \[L.H.L=R.H.L\] at \[x=1\], hence, limit of the function \[f\left( x \right)\] exists at \[x=1\] and the value of limit of the function \[f\left( x \right)\]at \[x=1\] is \[3\].

So, \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=3\]

Note: Graph of \[f\left( x \right)=\left\{ \begin{matrix}

x+2,x\ne 1 \\

0,x=1 \\

\end{matrix} \right.\] is given as

From the graph we can see, just slightly to the left of \[x=1\] and slightly to the right of \[x=1\], the value of \[f\left( x \right)\] is approximately equal to 3.

Hence, the value of \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=3\].

Also, the value of the limit will be equal to the value of the right hand limit and (or) the left hand limit.

We know, the limit of a function exists only if the left hand limit and right hand limit exist and both are equal.

So, first, we will find the left hand limit of the function \[f(x)\] at \[x=1\].

We know, the left hand limit of a function\[f\left( x \right)\] at the point \[x=a\] is given as

\[\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)\]

So , the left hand limit of the function\[f\left( x \right)\] at the point\[x=1\] is given as

\[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)\]

Now, we know h is a very small quantity and subtracting h from 1 will give a number less than 1 but very close to 1. But, since the number is not equal to 1, so, \[f\left( 1-h \right)=1-h+2\]

So, \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,1-h+2=1-0+2=3\]

Now, we will find the right hand limit of the function\[f\left( x \right)\] at the point \[x=1\].

We know right hand limit of a function \[f\left( x \right)\] at \[x=a\] is given by

\[R.H.L=\underset{h\to 0}{\mathop{\lim }}\, f\left( a+h \right)\]

So, the right hand limit of the function \[f\left( x \right)\] at the point \[x=1\] is given by

\[\underset{h\to 0}{\mathop{\lim }}\, f\left( 1+h \right)\]

Now, we know h is a very small number. So, 1+h will be slightly greater than 1 and not equal to 1. So, \[f\left( 1+h \right)=1+h+2\]

So, \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+h+2 \right)=3\]

We can clearly see that both the left hand limit and the right hand limit of the function \[f\left( x \right)\] exist at the point \[x=1\].

Also, the left hand limit and the right hand limit of the function \[f\left( x \right)\] at the point \[x=1\] are equal.

Since, the value of \[L.H.L=R.H.L\] at \[x=1\], hence, limit of the function \[f\left( x \right)\] exists at \[x=1\] and the value of limit of the function \[f\left( x \right)\]at \[x=1\] is \[3\].

So, \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=3\]

Note: Graph of \[f\left( x \right)=\left\{ \begin{matrix}

x+2,x\ne 1 \\

0,x=1 \\

\end{matrix} \right.\] is given as

From the graph we can see, just slightly to the left of \[x=1\] and slightly to the right of \[x=1\], the value of \[f\left( x \right)\] is approximately equal to 3.

Hence, the value of \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=3\].

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Which are the Top 10 Largest Countries of the World?

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

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

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

Select the word that is correctly spelled a Twelveth class 10 english CBSE