The proof of $\frac{d}{dx}\sin(x)$ goes something like this:

$$\begin{aligned} \lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\ =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}\\ =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\\ =0+\cos(x)\times1\\ =\cos(x) \end{aligned}$$

My doubt about this is that it uses the limit of $\frac{\sin(h)}h$ and $\frac{\cos(h)-1}{h}$ in the proof. However, proving these two limits uses L'Hopital's rule, which uses the derivative of $\sin(x)$ and $\cos(x)$ to prove the limits. This causes a circular argument because we're using the derivative of $\sin(x)$ to prove the derivative of $\sin(x)$. Is there a way to prove these two limits without using L'Hopital's rule or just looking at the graph, or is there a way to find $\frac{d}{dx}\sin(x)$ without using these two limits?

There is a nice way to prove the limits using geometry here, but I'm wondering if there's a way to do it without using this, either.

Edit: The way I'm defining sine is by the unit circle definition, not the Taylor series one.

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    $\begingroup$ I'm confused. You say that the limits use L'Hospital's rule, but then later you're aware of the geometric proofs that don't use any notions of derivatives. It's not circular for that reason.... $\endgroup$ – user296602 Sep 7 '17 at 17:07
  • $\begingroup$ @user296602 I'm asking if there's an algebraic way to prove the limit without L'Hopital's rule. $\endgroup$ – u8y7541 Sep 7 '17 at 17:07
  • $\begingroup$ Define the sine function purely algebraically, and we'll talk. (Note that if you define the sine function algebraically, you use Taylor series, and the limit is then very easy to prove.) $\endgroup$ – user296602 Sep 7 '17 at 17:08
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    $\begingroup$ Yea use the Taylor series for Sine. Edit: The Taylor series can actually be found without taking the derivative by looking at the Taylor series for $e^{ix}$ $\endgroup$ – FofX Sep 7 '17 at 17:11
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    $\begingroup$ @u8y7541 If you take the Taylor series as the definition of sine, then you have an analytical and algebraically useful definition (and this is how many authors do this). If you define it geometrically, it's natural that there's only a geometric proof. $\endgroup$ – user296602 Sep 7 '17 at 18:26

Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest, $\lim_{h\to 0}\frac{\sin(h)}{h}=1$, without appeal to geometry or differential calculus. (Note that $\cos(h)-1=-2\sin^2(h/2)$)

Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.

We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$$

for $|x|< 1$.

NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.

It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.

Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.

From $(1)$, we have the bounds (SEE HERE)

$$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$$

for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields

$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$

Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as

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

from which we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$$

as was to be shown!


We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.

Then, we have from $(2)$

$$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)} $$

which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.


This really comes down to how you define $\sin(x)$ and $\cos(x)$. As you say, there are geometric proofs that demonstrate the limits that you are asking about. You are correct that you can't use l'Hopital's rule to prove these limits.

A usual definition of $\sin(x)$ is through its Taylor series $$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots.$$ From here, you can see that $$\frac{\sin(h)}{h} = \frac{h - \frac{h^3}{6} + \frac{h^5}{120} - \cdots}{h} = 1 - \frac{h^2}{6} + \frac{h^4}{120} - \cdots \to 1$$ as $h\to 0$.

Similarly, it can be demonstrated that $\frac{\cos(x)-1}{h} \to 0$ as $h \to 0$.

You do not need to know what $\sin(x)$ to make this Taylor series. In Advanced Calculus courses, $\sin(x)$ is defined using the Taylor series, and later the trigonometric properties of $\sin(x)$ are demonstrated using the Taylor series as a starting point.

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    $\begingroup$ Moreover, Madhava of Sangamagrama discovered the series representations of sine, cosine, tangent and several other trigonometric functions more than a century before the invention of calculus. This demonstrates that the Taylor series for sine can be determined without knowing the derivatives of sine. $\endgroup$ – Joel Sep 7 '17 at 17:15
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    $\begingroup$ The arctan series was found via Riemann sum approach and then using the geometric series for $1/(1+x^{2})$. Notable here was the use of formula for sums of powers of first $n$ natural numbers. So this all boils down to integrating (via Riemann sum) the series $1-x^{2}+x^{4}-\cdots$. $\endgroup$ – Paramanand Singh Sep 7 '17 at 17:21
  • $\begingroup$ The series for since cosine was derived from that of arctan with some manipulation. $\endgroup$ – Paramanand Singh Sep 7 '17 at 17:22
  • $\begingroup$ Anyway the geometric definition of these functions are based on integrals so there is really no need of differentiation to obtain their properties. $\endgroup$ – Paramanand Singh Sep 7 '17 at 17:23
  • $\begingroup$ @ParamanandSingh Certainly, there are many ways to approach this problem. I like your idea of using integrals. $\endgroup$ – Joel Sep 7 '17 at 19:49

As mentioned in the comments above, it is all about your definition of $\sin$ and $\cos$. If you define it using Taylor series, then your proof probably ends up using that. If you use the "unit circle" to define the functions, then you proof will end up using some kind of geometry (since this is how you defined them).

As you note you don't have to use L'Hopital's rule to prove that $$ \lim_{h\to 0}\frac{\sin(h)}{n} = 1\quad\text{and}\quad\lim_{h\to 0}\frac{\cos(h) - 1}{h} = 0 $$ One can prove this limits from elementary geometric considerations.

It should be easy to find something using Google. I quickly found these notes.


This is too long for a comment.

The last sentence of the question gives the definition of sine being used here. To express the definition clearly, if one walks on the unit circle $x^{2}+y^{2}=1$ starting from point $A=(1, 0)$ in counter-clockwise direction and covers a distance $\theta $ (if $\theta$ is negative then the direction being clockwise) to reach point $P$ on the unit circle then the coordinates of this point $P$ are by definition $(\cos\theta, \sin\theta) $.

If the full circumference of unit circle is defined to be $2\pi$ then the length of a quadrant is $\pi/2$ and thus working only in the first quadrant we can see that that following formula holds $$\theta=\int_{0}^{\sin\theta}\frac{dt}{\sqrt{1 - t^{2}}}\tag{1}$$ for $0\leq\theta<\pi/2$. The above equation is the actual meaning of the geometric definition of sine function. Now we can argue like in user Mark Viola's answer.


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