Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (b)/Solution 1

First note that the function is defined at $x=1$ in the third piece of piecewise function, and $f(1)=1^2+1=2.$ When $k=2$, let’s see the left limit and right limit at point $x=1$ as follows: $$f(1^{+}):=\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }{x}^2+1=1^2+1=2$$ and $$f(1^{-}):=\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }k(x-1)+2=2.$$ We get that the left limit is equal to right limit, thus $$\underset{x\to 1}{\lim }f(x)$$ exists. And note that $$\underset{x\to 1}{\lim }f(x)=2=f(1)$$ which is the reason that $f(x)$ is continuous at $x=1$. So we choose $\text{(iv)}$.