Question

If
$\underset{x\to 2}{\lim }$$\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)

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