Science:Math Exam Resources/Courses/MATH102/December 2014/Question A 02

MATH102 December 2014

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Question A 02
What value of $a$ makes the function below continuous? Hint: Writing down the definition of the derivative of $\sin(x)$ at $x=0$ might be useful here. $f(x)=\left\{{\begin{matrix}{\tfrac {\sin(x)}{x}}&{\text{ if }}x\neq 0,\\a&{\text{ if }}x=0.\end{matrix}}\right.$ $a=0$ $a=1$ $a={\tfrac {1}{2}}$ $a={\tfrac {\sin(0)}{0}}$ $a={\tfrac {\pi }{2}}$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that a function $f$ is continuous at $x=p$ if $\lim _{x\to p}f(x)=f(p).$ What is the $p$ we must consider here, and what is the limit of $f$ as $x\to p?$

Hint 2
We observe that ${\tfrac {\sin(x)}{x}}$ is continuous for all $x\neq 0,$ so for $f$ to be continuous (everywhere), it suffices to have $\lim _{x\to 0}f(x)=f(0).$ What are $\lim _{x\to 0}f(x)$ and $f(0)?$

Hint 3
What is $\lim _{x\to 0}{\tfrac {\sin(x)}{x}}?$ To evaluate this limit, you may wish to recall that the derivative of a function $g(x)$ is by definition $g'(x)=\lim _{h\to 0}{\frac {g(x+h)-g(h)}{h}}.$ Alternatively, you could try using the squeeze theorem, l'Hôpital's rule, or Taylor series.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We observe that ${\tfrac {\sin(x)}{x}}$ is continuous for all $x\neq 0$ (as it is the quotient of continuous functions), so for $f$ to be continuous (everywhere), it suffices to have $\lim _{x\to 0}f(x)=f(0).$ Now $\lim _{x\to 0}f(x)=\lim _{x\to 0}{\tfrac {\sin(x)}{x}},$ while $f(0)=a.$ Hence for continuity we must have $a=\lim _{x\to 0}{\tfrac {\sin(x)}{x}}.$ This limit can be evaluated in numerous ways; we choose to follow the hint in the question statement and observe that by definition $\sin '(0)=\lim _{x\to 0}{\frac {\sin(0+x)-\sin(0)}{x}}=\lim _{x\to 0}{\frac {\sin(x)}{x}}$ since $\sin(0)=0.$ But we know that $\sin '(0)=\cos(0)=1,$ so the value of the limit is $1.$ Therefore we take ${\color {blue}{\text{(B) }}}a=1.$

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