Question

Consider the following function $g(x)=\left| x \right|,x\ne 0.$ Then,
I. $g(0)$ is not defined.
II. $\underset{x\to 0}{\mathop{\lim }}\,g(x)$ is not defined.
Which of the following is/are true?
A. Both I and II are true
B. Only I true
C. Only II is true
D. Both I and II are false

Answer 1

Hint : Do you remember the range of modulus of a function? First find the value of g(0), LHL and RHL. Then use the relation that the given function is continuous, when the left hand limit and the right hand limit should be equal to g(0). In this way you will be able to solve the question easily.

Complete step by step solution :
The given function is$g(x)=\left| x \right|,x\ne 0$
Now we will find the value of function at zero, so
$g(0)=|0|=0$
But this is not possible because it is given that it should not be defined at x=0, so $g(0)$ is not defined.
Now for the given function to be continuous, the left hand limit and the right hand limit should be equal to g(0), i.e., $g(0)=LHL=RHL$.
As the given function contains modulus, first we will separate the modulus with different range as shown below:

-x,x<0 \\\ x,x\ge 0 \\\ \end{matrix} \right.$$ First, we shall find the left hand limit, i.e., LHL. $$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,-x$$ Here we take $$x=0-h$$, where $$h\to 0$$then, $$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(x)=\underset{h\to 0}{\mathop{\lim }}\,-(0-h)$$ Applying the limits, we get $LHL=0........(i)$ Now, we shall find the right hand limit, i.e., RHL. $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x$$ Here we take $$x=0+h$$, where $$h\to 0$$then, $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g(x)=\underset{h\to 0}{\mathop{\lim }}\,(0+h)$$ Applying the limits, we get $$RHL=0........(ii)$$ From equation (i) and (ii), we see that $g(0)=LHL=RHL$ Hence, the given function is continuous at zero. So limit is defined at x=0. Hence, the correct answer is option (B). **Note** : Students forget to solve the modulus part of the function and they will make a mistake. When finding the left hand limit they forget to apply $$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(x)=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,g(0-h)$$this formula, instead they will think write $'0-h'$ instead of $'0+h'$ . In this way they will get different answers.