Science:Math Exam Resources/Courses/MATH105/April 2015/Question 01 (n)

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH105 April 2015

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q6 (a) • Q6 (b) •

Other MATH105 Exams

• April 2009 • April 2011 • April 2010 • April 2012 • April 2013 • April 2014 • April 2015 • April 2016 • April 2017 • April 2018 •

Question 01 (n)
Find the limit of the sequence $\left\{\ln \left(\sin {\frac {1}{n}}\right)+\ln(2n)\right\}$ .

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
To find the limit of the sequence $\{{a_{n}}\}$ , evaluate $\lim _{n\rightarrow \infty }a_{n}$

Hint 2
Recall that $\ln(x)+\ln(y)=\ln(xy)$

Hint 3
If you get stuck when evaluating a limit, try considering L'Hopital's rule.

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
To find the limit of the sequence $\left\{\ln \left(\sin {\frac {1}{n}}\right)+\ln(2n)\right\},$ we'll need to evaluate $\lim _{n\rightarrow \infty }\left[\ln \left(\sin {\frac {1}{n}}\right)+\ln(2n)\right].$ By the product rule for logarithms, $\ln(a)+\ln(b)=\ln(ab)$ , so this limit is equivalent to $\lim _{n\rightarrow \infty }\left[\ln \left(n\sin {\frac {1}{n}}\right)+\ln(2)\right]$ . We can begin by computing $\lim _{n\rightarrow \infty }\ln \left(n\sin {\frac {1}{n}}\right)=\ln \left(\lim _{n\rightarrow \infty }n\sin {\frac {1}{n}}\right)$ (by continuity of the logarithm). First, note that by using the substitution $u={\frac {1}{n}}$ , $\lim _{n\rightarrow \infty }n\sin {\frac {1}{n}}=\lim _{n\rightarrow \infty }{\frac {\sin {\frac {1}{n}}}{\frac {1}{n}}}=\lim _{u\rightarrow 0}{\frac {\sin u}{u}}.$ Now, we can apply L'Hôpital's rule for indeterminate forms (i.e., limits of the form ${\frac {0}{0}},{\frac {\infty }{\infty }}$ ) to compute $\lim _{u\rightarrow 0}{\frac {\sin u}{u}}$ , since substituting $u=0$ into ${\frac {\sin u}{u}}$ yields ${\frac {0}{0}}$ . L'Hôpital's Rule's rule gives $\lim _{u\rightarrow 0}{\frac {\sin u}{u}}=\lim _{u\rightarrow 0}{\frac {(\sin u)'}{u'}}=\lim _{u\rightarrow 0}{\frac {\cos u}{1}}=\cos(0)=1.$ Hence, $\ln \left(\lim _{n\rightarrow \infty }n\sin {\frac {1}{n}}\right)=\ln(1)=0.$ Therefore, $\lim _{n\rightarrow \infty }\left[\ln \left(\sin {\frac {1}{n}}\right)+\ln(2n)\right]=\lim _{n\rightarrow \infty }\left[\ln \left(n\sin {\frac {1}{n}}\right)+\ln(2)\right]=0+\ln(2)={\color {blue}{\ln(2)}}$ .