Question

The value of \[\;\mathop {\lim }\limits_{x \to 0} \left[ {\min ({y^2} - 4y + 11)\dfrac{{\sin x}}{x}} \right]\]where $ [.] $ denotes the greatest integer function) is,
A. $ 5 $
B. $ 6 $
C. $ 7 $
D. $ {\text{Does not exist}} $

Answer 1

Hint: We have to find the limit of the given function, the other important thing to note in this question here is that we are given the minimum function as well as the greatest integer function. We will first find the minimum value of the given quadratic function using differentiation and putting it to zero to find the critical points of the function and then checking if its maxima or minima at that point. Then we will solve the remaining part and after that solving the greatest integer function will our solution be completed in its entirety.
For finding minima we put $ y' = 0 $ for the function

Complete step-by-step answer:
We are given the limit,
\[\;\mathop {\lim }\limits_{x \to 0} \left[ {\min ({y^2} - 4y + 11)\dfrac{{\sin x}}{x}} \right]\]
First we will find the minimum value of the quadratic expression given in the question.
 $ f(y) = ({y^2} - 4y + 11) $
 $ f'(y) = 2y - 4 $
Putting first derivative equal to $ 0 $ , we get,
 $
  2y - 4 = 0 \\
  y = 2 \;
 $
At $ y = 2 $ , we check the second derivative ,
 $ f''(y) = 2 $ , which is positive and hence the function is minima at this point.
Thus the minimum value at $ 2 $ is
 $ f(2) = {2^2} + 11 - 4(2) $
 $ f(2) = 4 + 11 - 8 $
 $ f(2) = 7 $
Now we write our limit as,
\[\;\mathop {\lim }\limits_{x \to 0} \left[ {7\dfrac{{\sin x}}{x}} \right]\]
The number will be independent of the limit and we can write,
 $ \left[ {7.\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}} \right] $
Now we know that,
 $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1 $
Thus we get,
 $ [7 \times 1] $
 $ = 7 $
Hence the limit \[\;\mathop {\lim }\limits_{x \to 0} \left[ {\min ({y^2} - 4y + 11)\dfrac{{\sin x}}{x}} \right]\]is $ 7 $ which is the option C
So, the correct answer is “Option C”.

Note: Remember that even if any function gives a critical point at its first derivative, we can not know for sure that it will give a maxima or a minima, this means that only when we do the second derivative, we can tell, if the second derivative is positive it means it is minima , if the second derivative is negative it is a maxima also if it comes out to be $ 0 $ it is a inflexion point (neither maxima or minima)