JEE Main 2014 (Online) 11th AprilQuestion 1
If $\lim_{x\to2}\frac{\tan\left((x-2)\left(x^2+(k-2)x-2k\right)\right)}{x^2-4x+4}=5$, then $k$ is equal to :
Limits & Continuity
Solution
$x^2+(k-2)x-2k=(x-2)(x+k)$, so the argument is $(x-2)^2(x+k)$ and the limit is $2+k=5$, $k=3$.