Find the limit of $x(x + 1 - \sin(\frac{1}{1+x})^{-1})$ as $x \rightarrow \infty$ As the title states, I need to find the limit for $x\left(x + 1 - \frac{1}{\sin(\frac{1}{1+x})}\right)$ as $x \rightarrow \infty$, as part of a larger proof I am working on.
I believe the answer is 0. I think that to start, I can show that $\frac{1}{\sin(\frac{1}{1+x})} \rightarrow x + 1$ for large x. By looking at the series expansion for Sin, it's clear that Sin approximates to $\frac{1}{1+x}$ for large x, as the higher-power entries in the series $\frac{1}{1+x}^3 + \frac{1}{1+x}^5 + ...$ would disappear faster, but would it be sufficient to state this? Is there not a more rigorous way of showing this to be true?
If my approach is entirely wrong, or there is a more elegant way of reaching the answer, please share.
 A: Let $y=\dfrac{1}{x+1}$ then $y\to0$ in
\begin{align}
\lim_{xto\infty}x(x + 1 - \frac{1}{\sin(\frac{1}{1+x})})
&= \lim_{y\to0}\dfrac{(1-y)}{y}\left(\dfrac{1}{y}-\dfrac{1}{\sin y}\right)\\
&= \lim_{y\to0}\dfrac{(1-y)(\sin y-y)}{y^2\sin y}\\
&= \lim_{y\to0}\dfrac{(1-y)\left(y-\dfrac16y^3+O(y^5)-y\right)}{y^3}\\
&= \color{blue}{-\dfrac16}
\end{align}
A: According to Taylor's theorem with remainder, we have
$$\sin t=t-\frac{t^3}{6}+R_3(t) $$
where $|R_3(t)| \leq \frac{t^4}{24}$. Taking $t=\frac{1}{x+1}$ and plugging in your expression gives
$$\frac{x (x+1) \left(6 R_3(t) (x+1)^3-1\right)}{6 R_3(t) (x+1)^3+6 x (x+2)+5}=\frac{-x^2+o(x^2)}{6x^2+o(x^2)}  $$
which approaches $-1/6$ as $x \to \infty$.
A: To avoid Taylor's expansion by $y=\frac{1}{1+x}\to 0$ we have that
$$x\left(x + 1 - \frac1{\sin\left(\frac{1}{1+x}\right)}\right)=\frac{1-y}{y}\left(\frac1y - \frac1{\sin y}\right)=$$$$=\frac{1-y}{y}\left(\frac{\sin y -y}{y\sin y}\right)=(1-y)\frac{y}{\sin y}\left(\frac{\sin y -y}{y^3}\right)\to 1 \cdot 1 \cdot \left(-\frac16\right)=-\frac16$$
indeed we have that as $y\to 0$
$$\frac{\sin y -y}{y^3}\to -\frac16$$
refer to Are all limits solvable without L'Hôpital Rule or Series Expansion.
