Question

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 $\displaystyle \lim_{x \to 2}f\left( x \right)$
A. exists only when $K=-1$
B. exists for every real K
C. exists for every real K except $K=1$
D. does not exist

Answer 1

Hint: We first try to find the function and approaching value of the variable $x \to 2$. 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 $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=f\left( a \right)$. 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 $x=2$.
From the theorem we can tell that limit exists only when $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=f\left( a \right)$.
For our given limit the value of variable $x \to 2$. This means the value can be approaching from the both sides of the point of 2. We can break it into three parts of ${{2}^{+}},2,{{2}^{-}}$.
${{2}^{+}}$ represents that the value is approaching from the right-side or greater side of the point and ${{2}^{-}}$ 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 $\displaystyle \lim_{x \to {{2}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{2}^{-}}}f\left( x \right)=f\left( 2 \right)$.
$\displaystyle \lim_{x \to {{2}^{+}}}f\left( x \right)={{\left[ \dfrac{K\left( {{x}^{2}}-4 \right)}{2-x} \right]}_{x=2}}={{\left[ -K\left( x+2 \right) \right]}_{x=2}}=-K\left( 2+2 \right)=-4K$
$\displaystyle \lim_{x \to {{2}^{-}}}f\left( x \right)={{\left[ \dfrac{K\left( {{x}^{2}}-4 \right)}{2-x} \right]}_{x=2}}={{\left[ -K\left( x+2 \right) \right]}_{x=2}}=-K\left( 2+2 \right)=-4K$
$f\left( 2 \right)={{2}^{2}}=4$
Therefore, $-4K=4$ if the limit exists which gives that $K=\dfrac{4}{-4}=-1$.
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.