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

First, we note that the function is defined at ${\textstyle x=0}$ with ${\textstyle f(0)=e^{0}=1}$, so (i) and (iii) are not correct.

Now let’s see whether the limit of the function at $0$ exists. For this purpose, we consider the left limit and the right limit. When $x$ approaches to $0$ from the right, we consider points near $0$ which is greater than $0$. For such points, the corresponding piece of the function is $k(x-1)+2$, so that

$$\lim _{x\rightarrow 0^{+}}f(x)=\lim _{x\rightarrow 0^{+}}k(x-1)+2=-k+2=2$$

Here we use the given information $k=0$.

On the other hand, to get the left limit, we consider points $x<0$ close to $0$. Since the corresponding piece of the function in such region is $e^{x}$, we get

$$\lim _{x\rightarrow 0^{-}}f(x)=\lim _{x\rightarrow 0^{-}}e^{x}=e^{0}=1.$$

Since the left limit is not equal to right limit at ${\textstyle x=0}$,

$$\lim _{x\rightarrow 0^{+}}f(x)\neq \lim _{x\rightarrow 0^{-}}f(x).$$

the limit ${\textstyle \lim _{x\rightarrow 0}f(x)}$ does not exist. In all, we choose ${\textstyle \color {blue}{\text{(ii)}}}$.