# Proving that $\lim\limits_{x\to \infty}\,(\ln (2\pi x+\frac{\pi }{2})-\ln 2\pi x)=0$

Hello I am studying for an upcoming exam and I came to this question:

A) Prove that $\lim\limits_{x\to \infty}\,(\ln (2\pi x+\frac{\pi }{2})-\ln 2\pi x)=0$.

B) Use part A of the question to show that $\sin (e^x)$ is not uniformly continuous.

I am completely stumped - I do not see how to prove A or how it helps with B, Can someone please help me with this question? Thanks a lot :)

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Hint for a): $\ln a-\ln b=\ln{a\over b}$. Hint for b): What is $|\sin e^{\ln(2\pi x+\pi/2)} -\sin e^{\ln 2\pi x}|$ and what does a) tell you about $|\ln(2\pi x+\pi/2)-\ln 2\pi x|$? –  David Mitra Feb 15 '12 at 14:21
What does $\ln (2\pi > x + \frac{\pi}{2})$ mean? –  William Feb 15 '12 at 14:22
Ok, Thanks for the tips guys - I will try to solve using them :) –  Jason Feb 15 '12 at 14:23
OK, I see, that was a typo. –  William Feb 15 '12 at 14:25
@WNY It was a typo, I corrected it. –  Pedro Tamaroff Feb 15 '12 at 14:27

Hint for a):

Use the identity $\ln a−\ln b=\ln(a/b)$.

Hint for b):

What is $|\sin e^{\ln(2\pi x+\pi/2)}−\sin e^{\ln(2\pi x)}|$ and what does part a) tell you about $|\ln(2\pi x+\pi/2)−\ln(2\pi x)|$?

Warning: solution follows.

For part a), use the difference rules for logarithms: $$\ln(2\pi x+{\textstyle{\pi\over2}})-\ln(2\pi x)=\ln{ 2\pi x+{\pi\over2} \over 2\pi x }\quad \buildrel{x\rightarrow\infty}\over{\longrightarrow }\quad\ln (1)=0.$$

For part b), note that, for $x$ an integer $$\bigl|\,\sin e^{ \ln(2\pi x+{\textstyle{\pi\over2}})} -\sin e^{\ln 2\pi x}\,\bigr| =\bigl |\,\sin (2\pi x+{\textstyle{\pi\over2}}) -\sin (2\pi x)\,\bigr |=1.$$

Towards showing that $f(x)=\sin(e^x)$ is not uniformly continuous, let $\epsilon=1$ and suppose $\delta>0$.

By part a), we can select an integer $x$ so that $|\ln(2\pi x+{\textstyle{\pi\over2}})-\ln(2\pi x)|<\delta$.

But by part b), $$\bigl|\,\sin e^{ \ln(2\pi x+{\pi\over2})} -\sin e^{\ln( 2\pi x)}\,\bigr|=1= \epsilon.$$

So, we have demonstrated that for for $\epsilon=1$, and any $\delta>0$, we can find two points $x_1=\ln(2\pi x+{\pi\over2})$ and $x_2=\ln(2\pi x)$ such that $|x_1-x_2|<\delta$, yet $\bigl|\,\sin e^{x_1} -\sin e^{x_2}\,\bigr|\ge\epsilon$. This shows that $\sin(e^x)$ is not uniformly continuous on $\Bbb R$.

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Thanks for the great answer :) Have a nice day –  Jason Feb 15 '12 at 14:40