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Question: Let we have a function as \(f\left( x \right)=\left\\{ \begin{matrix} {{x}^{2}} & \text{ x is an ...

Let we have a function as f\left( x \right)=\left\\{ \begin{matrix} {{x}^{2}} & \text{ x is an integer} \\\ \dfrac{K\left( {{x}^{2}}-4 \right)}{2-x} & \text{ otherwise} \\\ \end{matrix} \right. then lim⁡x→2f(x)\displaystyle \lim_{x \to 2}f\left( x \right)
A. exists only when K=−1K=-1
B. exists for every real K
C. exists for every real K except K=1K=1
D. does not exist

Explanation

Solution

We first try to find the function and approaching value of the variable x→2x \to 2. Then we find the definition of limit and how it applies for the function to find the limit value. The limit only exists when the left-hand and right-hand each limit gives equal value. The mathematical form being lim⁡x→a+f(x)=lim⁡x→a−f(x)=f(a)\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=f\left( a \right). Using the formula, we find the value of K.

Complete step-by-step solution:
We assume the limit of the function f\left( x \right)=\left\\{ \begin{matrix} {{x}^{2}} & \text{ x is an integer} \\\ \dfrac{K\left( {{x}^{2}}-4 \right)}{2-x} & \text{ otherwise} \\\ \end{matrix} \right. exists at x=2x=2.
From the theorem we can tell that limit exists only when lim⁡x→a+f(x)=lim⁡x→a−f(x)=f(a)\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=f\left( a \right).
For our given limit the value of variable x→2x \to 2. This means the value can be approaching from the both sides of the point of 2. We can break it into three parts of 2+,2,2−{{2}^{+}},2,{{2}^{-}}.
2+{{2}^{+}} represents that the value is approaching from the right-side or greater side of the point and 2−{{2}^{-}} represents that the value is approaching from the left-side or lesser side of the point. There is also the fixed point of 2.
Now the limit value will exist only when lim⁡x→2+f(x)=lim⁡x→2−f(x)=f(2)\displaystyle \lim_{x \to {{2}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{2}^{-}}}f\left( x \right)=f\left( 2 \right).
lim⁡x→2+f(x)=[K(x2−4)2−x]x=2=[−K(x+2)]x=2=−K(2+2)=−4K\displaystyle \lim_{x \to {{2}^{+}}}f\left( x \right)={{\left[ \dfrac{K\left( {{x}^{2}}-4 \right)}{2-x} \right]}_{x=2}}={{\left[ -K\left( x+2 \right) \right]}_{x=2}}=-K\left( 2+2 \right)=-4K
lim⁡x→2−f(x)=[K(x2−4)2−x]x=2=[−K(x+2)]x=2=−K(2+2)=−4K\displaystyle \lim_{x \to {{2}^{-}}}f\left( x \right)={{\left[ \dfrac{K\left( {{x}^{2}}-4 \right)}{2-x} \right]}_{x=2}}={{\left[ -K\left( x+2 \right) \right]}_{x=2}}=-K\left( 2+2 \right)=-4K
f(2)=22=4f\left( 2 \right)={{2}^{2}}=4
Therefore, −4K=4-4K=4 if the limit exists which gives that K=4−4=−1K=\dfrac{4}{-4}=-1.
The correct option is A.

Note: The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.