Science:Math Exam Resources/Courses/MATH100 C/December 2024/Question 02 (c)

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MATH100 C December 2024

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Question 02 (c)
For this question, you do not need to justify your answers. Compute $\underset{x\to 0}{\lim }\frac{\arctan x}{x^3-6x}.$

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
We have that $\underset{x\to 0}{\lim }\arctan (x)=\underset{x\to 0}{\lim }{x}^3-6x=0$, so the limit is of the the indeterminate form $\frac{0}{0}$. What rule do you know that you can use to take limits of this form?

Checking a solution serves two purposes: helping you if, after having used the hint, 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

We can start by attempting to compute the limit using direct substitution as follows, $$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\arctan x}{x^3-6x}& =\frac{\underset{x\to 0}{\lim }\arctan x}{\underset{x\to 0}{\lim }{x}^3-6x}=\frac{0}{0}.\\ \end{array}$$ However this leads to an indeterminate form $\frac{0}{0}$, meaning we have to use L'Hopital's Rule instead: $$\begin{array}{rl}\underset{x\to a}{\lim }\frac{f(x)}{g(x)}& =\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)},\end{array}$$ provided the limit of the derivatives exists. Applying L'Hopital's Rule to the above limit, we get $$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\frac{1}{1+x^2}}{3x^2-6}& =\frac{\underset{x\to 0}{\lim }\frac{1}{1+x^2}}{\underset{x\to 0}{\lim }3x^2-6}=-\frac{1}{6.}\\ \end{array}$$

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