# Proving that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

So, I'm trying to prove the following assertion:

$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$

Proof Attempt:

Let $$\epsilon > 0$$. Then, we have to show that:

$$\exists \delta > 0 : 0 < |x| < \delta \implies |\frac{\sin(x)}{x} - 1| < \epsilon$$

Let us consider the following:

$$\cos(x) < \frac{\sin(x)}{x} < 1$$

if $$0 < |x| < \frac{\pi}{2}$$. So, we have:

$$0 < 1 - \frac{\sin(x)}{x} < 1 - \cos(x) = 2\sin^2(\frac{x}{2}) \leq 2 |\sin(\frac{x}{2})| \leq |x| < \delta$$

That's, of course, assuming that $$0 < |x| < \frac{\pi}{2}$$.

If $$\epsilon \geq \frac{\pi}{2}$$, then let $$\delta = \frac{\pi}{2}$$. This would ensure that $$1-\frac{\sin(x)}{x}$$ is still greater than 0.

If $$\epsilon < \frac{\pi}{2}$$, then let $$\delta = \epsilon$$. This would still mean that $$0 < |x| < \delta < \frac{\pi}{2}$$ so the inequality above would still be satisfied.

This proves the given assertion.

Could someone check my proof above and see if it works or not?

• – Invisible Mar 25 at 18:40
• If you solve the inequality, then $\delta$ works. (: – Invisible Mar 25 at 18:44
• Eyy nice, thanks for the link, I'll check out the discussion there as well. – Abhi Mar 25 at 18:45
• If you accept the "squeeze theorem," then letting $x\to 0$ in $$\cos(x)\le \frac{\sin(x)}{x}\le 1$$suffices. – Mark Viola Mar 25 at 19:16
• Can't accept that cos it hasn't been brought up in my text at this point, so I have to prove this in a more elementary way. Just to be sure, is there anything in my proof that could be improved upon? – Abhi Mar 25 at 19:18

First change the name of the variable to $$\theta,$$ so we have to prove $$\lim_{\theta\to 0}\frac{\sin \theta}{\theta}=1.$$ Take a point $$(x,y)$$ on the unit circle in the first quadrant. The proof is almost the the same for a point in the 4-th quadrant. Then $$\theta$$ is the arc length from $$(x,y)$$ to (1,0) and $$\sin \theta=y.$$ We need bounds on the arc length. For a non-negative integer $$n$$ divide the interval $$[x,1]$$ into $$2^n$$ equal sub-inervals. Note that we deliberately include the case $$n=0$$ when the interval is left in one piece. For each division point $$x_i$$ on the X-axis let $$P_i=(x_i,y_i), 0 \le i \le 2^n$$ be the corresponding point on the unit circle so that $$P_0=(x,y)$$ and $$P_{2^n}=(1,0).$$ For $$n \ge 0$$ let $$S_n=\sum_{j=1}^{2^n}\text {straight-line lengths from P_{j-1} to P_j}.$$ The triangle inequality shows that $$S_0,S_1, ...$$ is an increasing sequence and that $$S_n \le 1-x+y$$. Thus $$\lim_{n \to \infty}S_n$$ exists. We define the arc length=$$\theta=\lim_{n \to \infty}S_n$$Then $$\sqrt{(1-x)^2+y^2}=S_0 \le \theta \le 1-x+y$$. $$\sqrt{(1-\sqrt(1-y^2))^2+y^2}\le \theta \le 1-\sqrt(1-y^2)+y$$ The conditions $$\theta \to 0$$ and $$y \to 0$$ are equivalent. All that is needed now is to take the reciprocals in our upper and lower bounds for $$\theta$$, multiply through by $$\sin \theta$$, which is the same as $$y$$ to obtain upper and lower bounds on $$\frac {\sin \theta}{\theta}$$ in terms of $$y.$$ With the squeeze law for limits and a little algebra, you can finish the proof.

• Thank you for the argument. I can't use the squeeze law to prove this because it hasn't yet been proven in my text. This will be excellent reference for when I do learn it. – Abhi Mar 26 at 5:08

Let's prove an inequality using some geometry :

Let $$\theta \in\mathbb{R} \cdot$$ Observe the following figure : Denoting $$\mathscr{A}$$ the red region, (The triangle $$OBC$$), $$\mathscr{B}$$ the green one, (A part from the unit disc that has a circumference of $$\theta$$), $$\mathscr{C}$$ the ble one, (The triangle $$OCD$$). $$\mathscr{S}_{\mathscr{A}}$$, $$\mathscr{S}_{\mathscr{B}}$$ and $$\mathscr{S}_{\mathscr{C}}$$ are their respective areas.

We have : \begin{aligned} \mathscr{S}_{\mathscr{A}}\leq\mathscr{S}_{\mathscr{B}}\leq\mathscr{S}_{\mathscr{C}}\ \ \ \ \ \\ \iff \frac{\sin{\theta}\cos{\theta}}{2}\leq\frac{\theta}{2}\leq\frac{\tan{\theta}}{2}\ \ \ \\ \iff \ \ \ \ \ \ \ \cos{\theta}\leq\frac{\sin{\theta}}{\theta}\leq\frac{1}{\cos{\theta}}\end{aligned}

Meaning, we have $$\lim\limits_{\theta\to 0}{\frac{\sin{\theta}}{\theta}}=1 \cdot$$

• Thank you for the proof but I'm not quite allowed to use the squeeze theorem, which is what you utilized. We haven't proved that yet in the text I'm using. – Abhi Mar 26 at 5:06