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

First, we note that the function is defined at $x=0$ with $f(0)=e^0=1$, 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 $$\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }k(x-1)+2=-k+2=2$$ 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 $$\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }{e}^x=e^0=1.$$

Since the left limit is not equal to right limit at $x=0$, $$\underset{x\to 0^{+}}{\lim }f(x)\ne \underset{x\to 0^{-}}{\lim }f(x).$$ the limit $\underset{x\to 0}{\lim }f(x)$ does not exist. In all, we choose $\text{(ii)}$.