Question

If f\left( x \right)=\left\\{ \begin{aligned} & \sin \left[ x \right],\left[ x \right]\ne 0 \\\ & 0,\left[ x \right]=0 \\\ \end{aligned} \right.
where $\left[ x \right]$ denotes the greatest integer less than or equal to x. then
(a) $\displaystyle \lim_{x \to 0^-}f\left[ x \right]=\sin 1$
(b) $\displaystyle \lim_{x \to 0^+}f\left[ x \right]=0$
(c) limit does not exist at $x=0$
(d) limit exist at $x=0$

Answer 1

In this question, we first need to find the value of greatest integer function when x approaches 0 from the left hand side and the right hand side. Then we get the value of the function which gives the right hand limit and left hand limit. Now, if the right hand and left hand limits are equal then the limit exists.

Complete step-by-step answer:
Now, from the given function in the question we have

& \sin \left[ x \right],\left[ x \right]\ne 0 \\\ & 0,\left[ x \right]=0 \\\ \end{aligned} \right.$$ GREATEST INTEGER FUNCTION: For any real function the greatest integer function rounds-off the real number down to integer less than the number. This is denoted by $$\left[ x \right]$$. EXAMPLE: $$\begin{aligned} & \left[ 2.5 \right]=2 \\\ & \left[ -1.5 \right]=-2 \\\ \end{aligned}$$ LEFT HAND AND RIGHT HAND LIMITS: If the value of the function at the points which are very close to a on the left tends to a definite unique number as x tends to a, then that number is called the left hand limit. $$\displaystyle \lim_{x \to a^-}f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)$$ Similarly, right hand limit $$\displaystyle \lim_{x \to a^+}f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)$$ Existence of Limit: $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$$ exists, if $$\displaystyle \lim_{x \to a^-}f\left( x \right)=\displaystyle \lim_{x \to a^+}f\left( x \right)$$ Now, let us calculate the left hand limit for the given function As x reaches zero from the left hand side we get, $$x\to {{0}^{-}}$$ $$\left[ x \right]\to -1$$ As $$\left[ -0.9 \right]=-1$$ Now, the function becomes $$f\left( x \right)=\sin \left[ x \right]$$ Now, let us find the left hand limit value $$\Rightarrow \displaystyle \lim_{x \to 0^-}f\left[ x \right]=\displaystyle \lim_{x \to 0^-}\sin \left[ x \right]$$ Now, on substituting the corresponding value we get, $$\Rightarrow \displaystyle \lim_{x \to 0^-}f\left[ x \right]=\sin \left( -1 \right)$$ Now, on further simplification we get, $$\therefore \displaystyle \lim_{x \to 0^-}f\left[ x \right]=-\sin 1$$ Now, let us calculate the right hand limit of the given function As x reaches zero from the right hand side we get, $$x\to {{0}^{+}}$$ $$\left[ x \right]\to 0$$ As $$\left[ 0.1 \right]=0$$ Now, the function becomes $$f\left( x \right)=0$$ Now, let us find the right hand limit value $$\Rightarrow \displaystyle \lim_{x \to 0^+}f\left[ x \right]=\displaystyle \lim_{x \to 0^+}0$$ Now, on further simplification we get, $$\therefore \displaystyle \lim_{x \to 0^+}f\left[ x \right]=0$$ Here, we get that the right hand limit and left hand limit values are not equal. $$\therefore \displaystyle \lim_{x \to 0^-}f\left[ x \right]\ne \displaystyle \lim_{x \to 0^+}f\left[ x \right]$$ Thus, the limit of the given function does not exist at $$x=0$$ **So, the correct answers are “Option b and c”.** **Note:** It is important to note that the value of the left hand limit given in the question and the value we got are not equal because sine of a negative function is negative sine function not positive. Thus, this point should not be neglected as it changes the answer. It is also to be noted that the value of the greatest integer function when x approaches 0 from the left hand side will be -1 not 1 because when it approaches 0 from the left side the greatest integer less than will be -1 not 0.