Science:Math Exam Resources/Courses/MATH105/April 2015/Question 01 (n)/Solution 1

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,

$\lim _{n\rightarrow \infty }\left[\ln \left(\sin {\frac {1}{n}}\right)+\ln(2n)\right]=\lim _{n\rightarrow \infty }\left[\ln \left(n\sin {\frac {1}{n}}\right)+\ln(2)\right]=0+\ln(2)={\color {blue}{\ln(2)}}$.