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

MATH110 April 2016

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[hide] Question 03 (a)

Consider the following function, where $k$ is a constant. ${\begin{aligned}f(x)={\begin{cases}e^{x}&{\text{if }}x\leq 0\\k(x-1)+2&{\text{if }}0<x<1\\x^{2}+1&{\text{if }}x\geq 1.\end{cases}}\end{aligned}}$ For $k=0$, which one of the following statements is the best reason why $f(x)$ is not continuous at $x=0$? (i) $f(0)$ does not exist. (ii) $\lim _{x\to 0}f(x)$ does not exist. (iii) Both (i) and (ii). (iv) $f(0)$ and $\lim _{x\to 0}f(x)$ exist, but they are not equal.

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[show] Hint
Recall that a function is continuous at a point if 1, the function is defined at the point 2 the limit of the function exists at the point 3.the limit is equal to the value of the function at the point. If one of the conditions fails, we say that the function is discontinuous at the point.

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[show] Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\textstyle \color {blue}{\text{(ii)}}}$.