Question

If ${\lim }_{x\to 2}\frac{\tan (x-2)\cdot \left(x^2+(k-2)x-2k\right)}{\left(x^2-4x+4\right)}=5$, then the value of $k$ is ____

Answer 1

Answer: 3

We have $\underset{x\to 2}{\lim }\frac{\tan (x-2)\cdot \left(x^2+(k-2)x-2k\right)}{(x-2{)}^2}=5$
$\Rightarrow \underset{x\to 2}{\lim }\frac{\tan (x-2)}{x-2}\cdot \frac{\left(x^2+(k-2)x-2k\right)}{x-2}=5$
$\Rightarrow \underset{x\to 2}{\lim }\frac{x^2+(k-2)x-2k}{x-2}=5$
$\Rightarrow x^2+(k-2)x-2k=(x-2)(x+\lambda )$
For the value of limit to be $5,\lambda =3$.