Let $f(x)=\{\begin{array}{ll}kx+1& x\le 2\\ 3x-1& x>2\end{array}$. Find $k$ for continuity at $x=2$.

Solution

1. For the function to be continuous at $x=2$, the left-hand limit must equal the right-hand limit, so ${\lim }_{x\to 2^{-}}f(x)={\lim }_{x\to 2^{+}}f(x)$.
2. The left-hand limit is calculated as $k(2)+1=2k+1$.
3. The right-hand limit is calculated as $3(2) - 1 = 5$.
4. Setting the limits equal gives $2k + 1 = 5$, which simplifies to $2k = 4$ and $k=2$.
5. Hence the answer is (A).