Help proving the limit of this function (from the given definition) Definition: If, given any sequence $x_n \in D$ (where $D$ is a subset of the reals unbounded above), $\lim\limits_{x \to \infty} f(x)=L$ if, given any sequence $x_n \in D$ that diverges to $\infty$, $f(x_n) \to L$ (In the usual sense that, given any $\varepsilon>0$, there exists $N \in \Bbb{Z^+}$ such that for all $n \geq N, \vert x_n-L \vert <\varepsilon$) 
Claim: $\lim\limits_{x \to \infty} f(x)=\lim_{x\rightarrow \infty}\frac{x+\sin(x)}{x+1}=1.$
Proof
Let $x_n$ be a sequence in $\mathbb{R}-\{1\}$ for which $x_n \to \infty.$
We must show that $f(x_n) \to 1.$
Let $K \in \mathbb{R}$ be given, and let $\varepsilon>0$ be given.
Since $x_n \to \infty$, there exists $N\in \mathbb{Z^+}$ such that, for all $n \geq N, x_n>K.$
Now, for all $n \geq N:$
$\vert f(x_n)-1\vert=\vert \frac{x_n+\sin(x_n)}{x_n+1}\vert \leq \underbrace{\frac{\vert x_n \vert+\vert \sin(x_n) \vert}{\vert x_n+1 \vert}}_{\text{triangle inequality}}\leq\frac{\vert x_n \vert+1}{\vert x_n+1 \vert }<\underbrace{...}_{\text{what goes here?}}<\varepsilon$
Can someone (at least hint to) bridge the gap?
Thanks

$$
\frac{x}{x+1} + \frac{\sin x}{x+1}
$$
Given $\varepsilon>0$, find $N_1$ so large that whenver $x>N_1$, then $\dfrac{1}{x+1} <\dfrac\varepsilon2$.
Then $\left|\dfrac{\sin x}{x+1}\right|\le \left|\dfrac{1}{x+1}\right|<\dfrac\varepsilon2$.
Then finr $N_2$ so large that whenever $x>N_2$, then $\dfrac{x}{x+1}$ differs from $1$ by less than $\varepsilon/2$.
Then if $x>\max\{N_1,N_2\}$ then
$$
\left|\frac{x+\sin x}{x+1} - 1\right| \le \left|\frac{x}{x+1} - 1 \right| + \left| \frac{\sin x}{x+1} \right| < \frac\varepsilon2+\frac\varepsilon2.
$$
