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

MATH110 April 2016

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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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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