Question

If
$\displaystyle\lim_{x \to 2}$$\frac{tan \left(x-2\right)\left\{x^{2}+\left(k-2\right)x-2k\right\}}{x^{2}-4x+4} = 5,$
then k is equal to :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

$\displaystyle\lim_{x \to 2}\frac{tan \left(x-2\right)[x^{2}+kx-2k-2x)}{\left(x-2\right)^{2}} = 5$
$\displaystyle\lim_{x \to 2}\left(\frac{tan \left(x-2\right)}{\left(x-2\right)}\right)\frac{\left(x+k\right)\left(x-2\right)}{\left(x-2\right)} = 5$
$1. \left(2 + k\right) = 5$
$K = 3$