Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^2& x=2\\ \frac{k\left(x^2-4\right)}{2-x}& x\in Z-2\end{array}$ , then $\underset{x\to 2}{lim}f\left(x\right)$

• A. exists only when $k=-1$
• B. exits for every real k
• C. Exits for every real k except $k-1$
• D. does not exits

Answer 1

The correct option is A exists only when $k=-1$

Given $f\left(x\right)=\{\begin{array}{cc}{x}^2& x=2\\ \frac{k\left(x^2-4\right)}{2-x}& x\in Z-2\end{array}$

For limit to exist:

$\underset{x\to 2}{lim}{x}^2=\underset{x\to 2}{lim}\frac{k(x^2-4)}{2-x}$

$⟹4=\underset{x\to 2}{lim}\frac{k(x^2-4)}{2-x}$

Since above limit is of $\frac{0}{0}$ form, so using L'Hopital's rule:

$\underset{x\to 2}{lim}\frac{k(x^2-4)}{2-x}=\underset{x\to 2}{lim}\frac{k2x}{-1}=4$

$⟹-2k\times 2=4$

$⟹k=-1$