how to make natural logarithms very small Is the sequence of functions $f_n:\mathbb{R} \rightarrow \mathbb{R}$ defined by $$f_n(x)=\cos(n+x)+\log(1+\frac{\sin^2(n^nx)}{\sqrt{n+2}})$$ uniformly equicontinuous? Prove or disprove.
Note that $|f_n(x)-f_n(y)|\leq|\cos(n+x)-\cos(n+y)|+|\log\frac{1+\frac{sin^2(n^nx)}{\sqrt{n+2}}}{1+\frac{sin^2(n^ny)}{\sqrt{n+2}}}|$. I know the cosine function can become very small by taking $\delta = \epsilon$. But I have no idea whehter the natural logarithm can become very small or not. Can anyone give some hints?
 A: Maybe a slightly more detailed proof is needed. 
As you observed, the $\cos$ are not annoying. It is just enough to show that the sequence $(g_n)$ is unformly equicontinuous, where 
$$g_n(x)=\log\left( 1+\frac{\sin^2(n^nx)}{\sqrt{n+2}}\right)\cdot $$
It is convenient to set $$\alpha_n(x):=1+\frac{\sin^2(n^nx)}{\sqrt{n+2}}\cdot$$ Since $1\leq \alpha_n(x)\leq \frac1{\sqrt{n+2}}$ for all $x\in\mathbb R$, we see that $\alpha_n(x)\to 1$ uniformly on $\mathbb R$. Hence, $g_n(x)=\log(\alpha_n(x))\to 0$ uniformly on $\mathbb R$.
Moreover (and this is where I think some more details were needed), for each fixed $n\in\mathbb N$ the function $g_n$ is uniformly continuous on $\mathbb R$. To see this, observe that the function $\alpha_n$ is uniformly continuous (being continuous and periodic), and that $1\leq \alpha_n(x)\leq 2$ for all $x\in\mathbb R$. Since $\log$ is uniformly continuous on the compact interval $[1,2]$, it follows that $g_n=\log(\alpha_n)$ is indeed uniformly continuous.
Now, you should have no difficulty in showing that the sequence $(g_n)$ is uniformly equicontinuous: given $\varepsilon >0$, start by choosing $N$ such that $\Vert g_n\Vert_\infty <\varepsilon/3$ for all $n> N$ (so that $\vert g_n(x)-g_n(y)\vert<2\varepsilon/3$ for any $x,y\in\mathbb R$ if $n> N$); and then use the uniform continuity of $g_1,\dots ,g_N$.
Note that everything works in the same way if the function $\log$ is replaced by any function $\phi$ which is continuous on $(0,\infty)$ and such that $\phi(1)=0$.
A: If a >= 1 and b >= 1 then |log a - log b| < |a - b)|.
That's because the derivative of log x is < 1 for all x > 1.
Also |cos a - cos b| < |a - b|.
So the answer is YES.
