# As a final illustration, find the value of \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)\] where

\[f\left( x \right)=\left\{ \begin{matrix}

x+2,x\ne 1 \\

0,x=1 \\

\end{matrix} \right.\]

Answer

Verified

363.3k+ 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 \[\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=3\].

Last updated date: 25th Sep 2023

â€¢

Total views: 363.3k

â€¢

Views today: 8.63k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

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

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers