Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
2  
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
show 1 more comment

1 Answer

up vote 7 down vote accepted

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 $\Bbb R$.

share|improve this answer
1  
Thanks for the great answer :) Have a nice day –  Jason Feb 15 '12 at 14:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.