$\lim_{x \to \infty} f(x)=1 $ $\implies$ $f(x) \sin x$ is uniformly continuous on $\mathbb R$? Let $f:\mathbb R \to \mathbb R$ be a continuos function such that $\lim_{x \to \infty} f(x)=1 $ , then is it true that $f(x) \sin x$ is uniformly continuous on $\mathbb R$ ? 
 A: Assuming that you add the hypothesis that $\lim_{x \rightarrow -\infty} f(x) = 1$, then it is true.
Let $\epsilon > 0$. Then there exists some positive $N$ such that for all $|x| > N$ we have $|f(x) - 1| < \epsilon$.
Since $\sin$ is uniformly continuous, there exists a $\delta > 0$ such that whenever $|x-y| < \delta$, we have $|\sin(x) - \sin(y)| < \epsilon$. Therefore, if $|x-y| < \delta$ and both $|x|$ and $|y|$ are larger than $N$, 
$$\begin{align}
|f(x)\sin(x) - f(y)\sin(y)| &=
|f(x)\sin(x) - f(x)\sin(y) + f(x)\sin(y) - f(y)\sin(y)| \\
&\leq |f(x)\sin(x) - f(x)\sin(y)| + |f(x)\sin(y) - f(y)\sin(y)| \\
&= |f(x)||\sin(x) - \sin(y)| + |\sin(y)| |f(x) - f(y)| \\
&\leq \epsilon|f(x)| + |f(x)- 1 + 1 - f(y)| \\
&\leq \epsilon(1 + \epsilon) + |f(x) - 1| + |f(y) - 1| \\
&\leq \epsilon(1 + \epsilon) + 2\epsilon
\end{align}$$
Obviously we use a smaller $\delta$ if needed, to make the right hand side equal to $\epsilon$.
So that takes care of $|x| > N$.
Now the interval $|x| \leq N$ is compact, and $f(x)\sin(x)$ is continuous, so it is uniformly continuous there. Thus there is some $\delta'$ such that $|f(x)\sin(x) - f(y)\sin(y)| < \epsilon$ whenever $|x-y| < \delta'$ and $|x|$ and $|y|$ are both $\leq N$.
Taking the smaller of $\delta$ and $\delta'$ gives us what we want, except we didn't handle the annoying case where $|x| \leq N$ and $|y| > N$ or vice versa. I'll leave you to patch up that bit: instead of looking at the interval $|x| \leq N$, we could have taken $|x| \leq N + 1$, and require both $\delta$ and $\delta'$ to be smaller than $1/2$.
A: The answer is yes. You can easily get the conclusion if you have learned two facts below.


*

*Let $g\in C(a,b)$. If $\lim\limits_{x\to a}g$ and $\lim\limits_{x\to b}g$ exist, then $g$ is uniformly continuous in $(a,b)$. That is to say, $f$ is uniformly continuous in $\mathbb R$.

*$\sin x$ is uniformly continuous in $\mathbb R$.

Estimate the value
$$|f(x)\sin x-f(y)\sin y|$$
$$\leq|f(x)||\sin x-\sin y|+|\sin y||f(x)-f(y)|$$
$$\leq M|\sin x-\sin y|+|f(x)-f(y)|$$
which will lead to the answer.
