Evaluate the following limit:
$\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}\left[ \cos x \right]}{1+\left[ \cos x \right]}$ ($\left[ . \right]$ denotes the greatest integer function)
(a) 0
(b) -1
(c) 1
(d) $\infty $

Hint: To evaluate the above limit we have to first solve the greatest integer function. We are going to solve the value of $\left[ \cos x \right]$. As x is tending to 0 so the value of $\cos x$ is tending to 1. Now, the greatest integer function when operated on the positive numbers gives the integer just less than that number. For e.g., when we apply greatest integer function on the positive number say 5.6 then the integer which is just less than 5.6 is 5. Then we will simplify the fraction given above.

Complete step by step answer:
The limit expression given in the above problem is as follows:
$\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}\left[ \cos x \right]}{1+\left[ \cos x \right]}$
Now, we are going to solve the greatest integer function of $\cos x$ after applying the limit on $\left[ \cos x \right]$. It is given above that $''x''$ is tending to 0 so the angle of the cosine is also tending to 0. And we know that the value of $\cos 0=1$. This means the value of $\cos x$ is tending to 1. Now, the value of $\cos x$ is lying between 0 and 1 so application of the greatest integer function on the number lying between 0 and 1 is 0.
Now, the fraction given above is reduced to:
$\dfrac{{{\sin }^{-1}}\left( 0 \right)}{1+0}$
We know that ${{\sin }^{-1}}\left( 0 \right)=0$ so using this relation in the above fraction we get,
$\begin{align}
  & \dfrac{0}{1} \\
 & =0 \\
\end{align}$
From the above, the evaluation of the above limit is equal to:
$\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}\left[ \cos x \right]}{1+\left[ \cos x \right]}=0$

Hence, the correct option is (a).

Note:
The mistake that could happen in the above problem is that you might write the greatest integer function of $\cos x$ tending to 1 as 1 because you think that as the value of $\cos x$ is tending to 1 so the value of $\left[ \cos x \right]$ is also 1. This is wrong because the value of $\cos x$ is tending to 1 not equal to 1. When the value of $\cos x$ is tending to 1, it means the value of $\cos x$ is lying between 0 and 1 and we know that the greatest integer function application on the number lying between 0 and 1 is 0.
So, make sure you won’t make this mistake in the examination.