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

MATH110 April 2016

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[hide] Question 03 (b)

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 $k=2$, which one of the following statements is the best reason why $f(x)$ is continuous at $x=1$? (i) $f(1)$ exists. (ii) $\lim _{x\to 1}f(x)$ exists. (iii) Both (i) and (ii). (iv) $f(1)$ and $\lim _{x\to 1}f(x)$ exist, and they are equal.

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[show] Hint
Recall the definition of the continuity of a function at a point (See the Hint of the part (a) in this question.) Note that all three conditions should be satisfied to be continuous.

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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 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 ${\textstyle \color {blue}{\text{(iv)}}}$.