Proving $\lim\limits_{u \to 0} \frac{\sin 2u}{2u} = 1$ without L'Hospital's rule

Question

I am just trying to see whether this below proof is valid. I want to prove that $$ \lim\limits_{u \to 0} \frac{\sin 2u}{2u} = 1. $$ I am taking for granted that $\lim\limits_{x \to 0} \frac{\sin x}{x}$, the proof of which I am fine with.

Let $t = 2u$. Then $t \to 0$ if and only if $u \to 0$. So $$ \lim\limits_{u \to 0} \frac{\sin 2u}{2u} = \lim\limits_{t \to 0} \frac{\sin t}{t} = 1. $$

Does this look ok? The key step is the claim, which may be false, that $t \to 0$ if and only if $2u \to 0$. I believe this follows from the fact that the map $t = 2u$ is a bijection, but I'm not totally sure how to prove it rigorously.

Answer 1

Let $N, M, K$ be metric spaces. Suppose $f : N \to M$ is a function, $x \in N$, and for all $y \in N$, $f(y) = f(x)$ iff $y = x$. Suppose further that $f$ is continuous at $x$.

Suppose $\lim\limits_{z \to f(x)} g(z)$ exists, $g : M \setminus \{f(x)\} \to K$.

Then $\lim\limits_{y \to x} g(f(x)) = \lim\limits_{z \to f(x)} g(z)$.

The proof is a straightforward $\delta-\epsilon$ proof.

Edit: the proof is as follows. Let $L = \lim\limits_{z \to f(x)} g(z)$. Consider some $\epsilon > 0$. Take some $\delta_1 > 0$ such that for all $z \in M$ s.t. $0 < |z - f(x)| < \delta_1$, $|g(z) - L| < \epsilon$. Take $\delta_2 > 0$ such that for all $y \in N$, $0 < |y - x| < \delta_2$ implies $|f(y) - f(x)| < \delta_1$. Then for all $y \in N$ such that $0 < |y - x| < \delta_2$, we have $0 < |f(y) - f(x)| < \delta_1$ and therefore $|g(f(y)) - L| < \epsilon$.

We apply the above theorem. In this case, we have $f(x) = 2x$, $g(x) = \frac{\sin x}{x}$.

Answer 2

Since I don't feel like getting into technicalities about formulating precisely a theorem about limits of not-necessarily continuous compositions, I'll deal with the continuous case.

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Method 1.

Define $\sigma:\Bbb{R}\to\Bbb{R}$ as \begin{align} \sigma(x):= \begin{cases} \frac{\sin x}{x}&\text{if $x\neq 0$}\\ 1 & \text{if $x=0$} \end{cases} \end{align}

Since you're assuming $\frac{\sin x}{x}\to 1$ as $x\to 0$, it follows $\sigma$ is continuous (on all of $\Bbb{R}$, even at the origin). Next, consider the "doubling function" $\delta:\Bbb{R}\to\Bbb{R}$, given as $\delta(u):=2u$. Then, $\sigma\circ \delta$ is the composition of continuous functions thus continuous. Also, from the definitions, \begin{align} (\sigma\circ \delta)(u)&=\sigma(2u)\\ &=\begin{cases} \frac{\sin 2u}{2u}&\text{if $2u\neq 0$}\\\\ 1 & \text{if $2u=0$} \end{cases}\\\\ &= \begin{cases} \frac{\sin 2u}{2u}&\text{if $u\neq 0$}\\\\ 1 & \text{if $u=0$} \end{cases} \end{align} Therefore \begin{align} \lim\limits_{u\to 0}\frac{\sin 2u}{2u}&=\lim\limits_{u\to 0}(\sigma\circ \delta)(u)=(\sigma\circ \delta)(0)=\sigma(\delta(0))=\sigma(0)=1. \end{align} (the second equality used continuity of the composition).

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Method 2.

We can just use the $\epsilon,\delta$ definition. Let $\epsilon>0$ be arbitrary. Since $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$, by definition there exists a $\delta>0$ such that if $x\in\Bbb{R}$ is any real number such that $0<|x|<\delta$, then \begin{align} \left|\frac{\sin x}{x}-1\right|<\delta. \end{align}

Now, choose $\delta':=\frac{\delta}{2}$, and let $u\in\Bbb{R}$ be any real number satisfying $0<|u|<\delta'$. Then, multiplying by $2$, we see $2u$ is a real number such that $0<|2u|<\delta$. Therefore, \begin{align} \left|\frac{\sin 2u}{2u}-1\right|<\epsilon. \end{align}

Since $\epsilon>0$ was arbitrary, this completes the proof that $\lim\limits_{u\to 0}\frac{\sin 2u}{2u}=1$.

Just to be clear, I showed that

For every $\epsilon>0$, there exists a $\delta'>0$ such that for all $u\in\Bbb{R}$, if $0<|u|<\delta'$, then $\left|\frac{\sin 2u}{2u}-1\right|<\epsilon$,

which is precisely the definition of limits $\lim\limits_{u\to 0}\frac{\sin 2u}{2u}=1$. In the course of the proof, I had to use the assumption $\frac{\sin x}{x}\to 0$ to first, given $\epsilon>0$, get a $\delta>0$ and only after that get a $\delta'>0$ (this is a very common idea, and pretty much how "composition of continuous functions is continuous" is proven, so it's good to get used to such reasoning).