Evaluate $\lim_{x\to+\infty} \frac{x-\sin{x}}{x+\sin{x}}$

Question

Evaluate $$\lim_{x\to+\infty} \frac{x-\sin{x}}{x+\sin{x}}$$

$\sin{x}$ is a bounded function, but I still have no clue what to do. Any hint?

Answer 1

Use:

$$ \frac{x-\sin x}{x+\sin x}=\frac{x+\sin x-2\sin x}{x+\sin x}=1-\frac{2\sin x}{x+\sin x} $$

It should be easy enough to evaluate the limit using the rightmost expression.

Answer 2

The bounds $\frac{x\mp1}{x\pm1}=\left(\frac{1+1/x}{1-1/x}\right)^{\mp1}$ provide an easy use of the squeeze theorem.

Answer 3

Have you ever heard about the squeeze theorem? Here I present its statement and the corresponding proof.

Let $(X,d_{X})$ be a metric space, $E\subseteq X$, $x_{0}\in X$ such that $x_{0}$ is an adherent point of $E$ and $L\in \textbf{R}$. Moroever, let us consider that $f:X\to\textbf{R}$, $g:X\to\textbf{R}$ and $h:X\to\textbf{R}$ are real-valued functions s.t. \begin{align*} f(x)\leq g(x) \leq h(x)\quad\wedge\quad\lim_{x\to x_{0};x\in E}f(x) = \lim_{x\to x_{0};x\in E}h(x) = L \end{align*} Thus we can conclude that $\displaystyle\lim_{x\to x_{0};x\in E}g(x) = L$.

Proof

Let $\varepsilon > 0$. Then there exists $\delta_{1} > 0$ and $\delta_{2} > 0$ s.t. for every $x\in E$ the next implications hold \begin{align*} \begin{cases} d_{X}(x,x_{0}) < \delta_{1} \Rightarrow |f(x) - L| < \varepsilon\\\\ d_{X}(x,x_{0}) < \delta_{2} \Rightarrow |h(x) - L| < \varepsilon \end{cases} \end{align*}

Based on the previous results, we may claim that for every $\varepsilon > 0$, there exists a $\delta = \min\{\delta_{1},\delta_{2}\}$ s.t. \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon \Rightarrow |g(x) - L| < \varepsilon \end{align*} and we conclude that $\displaystyle\lim_{x\to x_{0};x\in E}g(x) = L$.

At your case, we are taking limits as $x$ approach $+\infty$, but the same reasoning applies. It suffices to consider the following definition: \begin{align*} \lim_{x\to\infty}g(x) = L \Longleftrightarrow (\forall\varepsilon > 0)(\exists N\in\textbf{R}_{\geq0})\,\,\text{s.t.}\,\,x > N \Rightarrow |g(x) - L| < \varepsilon \end{align*}

To solve the proposed exercise, notice that \begin{align*} \frac{x - \sin(x)}{x + \sin(x)} = \frac{1 - \frac{\sin(x)}{x}}{1 + \frac{\sin(x)}{x}} \end{align*}

Since $|\sin(x)|\leq 1$ and $1/x\to 0$ as $x\to\infty$, we conclude that \begin{align*} \lim_{x\to\infty}\frac{x-\sin(x)}{x+\sin(x)} = 1 \end{align*}

Answer 4

What if $x=1\text{ trillion?}$ Then we have $$ \frac{1\text{ trillion} - \text{a tiny amount}}{1 \text{ trillion} + \text{the same tiny amount}} \approx 1. $$ The approximation can be made as close to the truth as desired by putting a big enough number in place of $1\text{ trillion}.$ So the limit is $1.$