2

Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not .

Answer:

Since $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})]=0 , \ \ and \ \ - 1\leq \sin(2x) \leq 1 $, the given limit oscillates between $ -1 \ \ to \ \ 1 $.

So the limit does not exists .

I need confirmation about my work. Any help is there ?

MAS
  • 10,638
  • 6
    You are correct. – vadim123 Jul 10 '17 at 04:06
  • 7
    You should (a) be more precise about what $-1\leq \sin(2x) \leq 1$ means - more specifically, that it attains both endpoints infinitely often as $x\to\infty$ - and (b) be more precise about why these two things mean that the limit can't exist. But your intuition is correct. – Steven Stadnicki Jul 10 '17 at 04:20

1 Answers1

0

Let us consider the following two sequences: $$a_n=n\pi+\frac{\pi}{4}$$ $$b_n=n\pi+\frac{3\pi}{4}$$

and let $f(x)=\ln\left(1+\frac{1}{x}\right)+\sin(2x)$

So, we see that: $$\begin{align*}f(a_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(\frac{\pi}{2}\right)=\\=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+1\overset{n\to\infty}{\longrightarrow}1 \end{align*}$$ and $$\begin{align*}f(b_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{3\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(\frac{3\pi}{2}\right)=\\=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)-1\overset{n\to\infty}{\longrightarrow}-1 \end{align*}$$ So, the limit $\lim\limits_{x\to+\infty}f(x)$ does not exist since we have shown that $f$ has at least two accumulation points as $x\to+\infty$.

Vassilis Markos
  • 4,484