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

First note that the function is defined at ${\textstyle x=1}$ in the third piece of piecewise function, and ${\textstyle f(1)=1^{2}+1=2.}$ When ${\textstyle k=2}$, let’s see the left limit and right limit at point ${\textstyle x=1}$ as follows:

$$f(1^{+}):=\lim _{x\rightarrow 1^{+}}f(x)=\lim _{x\rightarrow 1^{+}}x^{2}+1=1^{2}+1=2$$

and

$$f(1^{-}):=\lim _{x\rightarrow 1^{-}}f(x)=\lim _{x\rightarrow 1^{-}}k(x-1)+2=2.$$

We get that the left limit is equal to right limit, thus

$$\lim _{x\rightarrow 1}f(x)$$

exists. And note that

$$\lim _{x\rightarrow 1}f(x)=2=f(1)$$

which is the reason that ${\textstyle f(x)}$ is continuous at ${\textstyle x=1}$. So we choose ${\textstyle \color {blue}{\text{(iv)}}}$.