Question

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

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is C 3

$\underset{x\to 2}{lim}\frac{tan(x-2)[x^2+(k-2)x-2k]}{x^2-4x+4}=5$

put $x=2+h$

$\underset{h\to 0}{lim}\frac{tanh\cdot [h^2+(k+2\left)h\right]}{h^2}$

$\underset{h\to 0}{lim}\left[\frac{tanh}{h}\right]\cdot [h+k+2]=5$

$\Rightarrow 1.(0+k+2)=5$

$\Rightarrow k=3$