I'm working on an $\epsilon$-$\delta$ proof for $$\lim_{x\to \pi/2} \sin x=1$$

Here are the steps I've taken so far up to where I'm stuck:

Iff $\forall \epsilon>0$, $\exists \space \delta>0 \space s.t. \lvert sin(x)-1\rvert<\epsilon$ for any $0<\lvert x-\pi/2\rvert<\delta$

Take $\epsilon>0$

$\lvert sin(x)-1\rvert<\epsilon$

$\implies 1-\epsilon<\sin x<1+\epsilon$

Here's where I'm not sure how to proceed. I want to isolate $x$, so my initial inclination is to take the arcsin:

$\implies \arcsin (1-\epsilon)<x<\arcsin (1+\epsilon)$

But I know that arcsin $(1+\epsilon)$ is not valid, given that the domain can't exceed 1, so I'm not sure how to continue from here.

  • 1
    $\begingroup$ How do you define $\sin x$? $\endgroup$ – Hagen von Eitzen Sep 6 '16 at 21:21

You have $$ \sin x-1=\sin x-\sin\frac{\pi}{2}= 2\sin\frac{x-(\pi/2)}{2}\cos\frac{x+(\pi/2)}{2} $$ so $$ \lvert\sin x-1\rvert\le 2\left|\sin\frac{x-(\pi/2)}{2}\right| $$ because $\lvert\cos t\rvert\le 1$.

Hence, as soon as $\lvert\sin(x/2-\pi/4)\rvert<\varepsilon/2$ you surely have $\lvert\sin x-1\rvert<\varepsilon$.

Now a simple geometric construction (the same you use for proving that $\lim_{x\to0}\frac{\sin x}{x}=1$) shows that $\lvert\sin t\rvert\le\lvert t\rvert$ so, as soon as $\lvert x/2-\pi/4\rvert<\varepsilon/2$, that is, $$ \left|x-\frac{\pi}{2}\right|<\varepsilon $$ you have the required inequality.

More generally, if you want to prove $\sin x$ is continuous at $x$, you can do in the same way. Given $\varepsilon>0$, we want to find $\delta>0$ such that, if $|h|<\delta$, $\lvert\sin(x+h)-\sin x\rvert<\varepsilon$.

Suppose $|h|<\varepsilon$; then $$ \varepsilon>2\left|\frac{h}{2}\right|\ge 2\left|\sin\frac{h}{2}\right|\ge \left|2\sin\frac{h}{2}\cos\left(x+\frac{h}{2}\right)\right|= \lvert\sin(x+h)-\sin x\rvert $$ using the same sum-to-product formula as before. The basic fact is again that $\lvert\sin t\rvert\le|t|$ which by itself proves the sine is continuous at $0$.

  • $\begingroup$ (+1) This avoids using the MVT or Taylor series. $\endgroup$ – user84413 Sep 6 '16 at 23:13
  • $\begingroup$ Thanks! That helped a lot:) $\endgroup$ – Charlotte Sep 10 '16 at 22:41

You don't take $\epsilon$, you are given it. Then you have to come up with a $\delta$ that works. So you need to have a $\delta$ such that for any $x \in (\frac \pi 2-\delta, \frac \pi 2+ \delta), 1-\epsilon \lt \sin x \lt 1+\epsilon$. As $\sin x \le 1$ the right inequality is no problem. The easiest approach is to expand $\sin x$ in a Taylor series around $\frac \pi 2$. The series is alternating, so you know the error is of the sign of the first neglected term and smaller than it.

  • 2
    $\begingroup$ You are using Taylor expansion to prove a statement about continuity. If we know that $\sin$ is differentiable at $\pi/2$, why bother check continuity? $\endgroup$ – Banach Manifold Sep 6 '16 at 21:27


You can use the Mean Value Theorem to get that

$\sin x-\sin \frac{\pi}{2}=(\cos c)(x-\frac{\pi}{2})$ for some $c$ between $x$ and $\frac{\pi}{2}$,

so $\big|\sin x-1\big|=\big|\cos c\big|\big|x-\frac{\pi}{2}\big|\le\big|x-\frac{\pi}{2}\big|$.

(Notice that this argument could be considered circular, since it assumes differentiability of the sine function, but it does give a modulus of continuity.)


Using the identity $\sin a - \sin b = 2\cos((a+b)/2)\sin((a-b)/2$, we have $$\mid \sin x -1 \mid =\mid \sin x - \sin \pi/2 \mid = \mid 2\cos(x/2+\pi/4)\sin(x/2-\pi/4) \mid \leq 2\mid\sin(x/2 - \pi/4)\mid.$$

Usually it is argued geometrically that for $\theta$ near $0$, $0\leq \sin\theta \leq \theta$. So let $\epsilon >0$ be given. Let $\delta = \epsilon/2$ and suppose $\mid x -\pi/2 \mid <\delta = \epsilon/2$. Then $$\mid \sin(x) -1 \mid \leq 2\mid\sin(x/2 - \pi/4)\mid \leq 2\mid x/2 -\pi/4 \mid <2\delta =\epsilon.$$

And we've avoided the circular reasoning.


When $f$ is an increasing function, you can use the approach $a < f(x) < b \impliedby f^{-1}(a) < x < f^{-1}(b)$. But $\sin$ is not an increasing function in the neighborhood of $\frac\pi2$ (rather, it reaches a maximum there and is less on either side of $\frac\pi2$), so this approach does not work.

The function $\arcsin$ is defined so that $\arcsin y$ gives exactly one value of $x$ such that $\sin x = y$. Fortunately for you, the value it gives is exactly one of the values you want: $$\sin(\arcsin(1-\epsilon)) = 1 - \epsilon \quad\text{if}\quad 0 \leq \epsilon \leq 2.$$ Note: technically, you need to provide a different way to obtain $\delta$ in the case where $\epsilon > 2$, but it should not be hard to find a suitable $\delta$ in that case.

To find the upper bound for $x$, consider the fact that the $\sin$ function is symmetric around $\frac\pi2$, that is, $\sin\left(\frac\pi2 - \theta\right) = \sin\left(\frac\pi2 + \theta\right)$.

It may make things easier if you use $\arccos$ instead, remembering that $$ \arccos y = \frac\pi2 - \arcsin y \quad\text{if}\quad -1 \leq y \leq 1.$$


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