I have heard people say that you can't (or shouldn't) use the L'Hopital's rule to calculate $\lim\limits_{x\to 0}\frac{\sin x}x=\lim\limits_{x\to 0}\cos x=1$, because the result $\frac d{dx}\sin x=\cos x$ is derived by using that limit.

But is that opinion justified? Why should I be vary of applying L'Hopital's rule to that limit?

I don't see any problem with it. The sine function fulfills the conditions of the L'Hopital's rule.

Also, it is a fact that the derivative of sine is cosine, no matter how we proved it. Certainly there is a way to prove $\frac d{dx}\sin x=\cos x$ without using the said limit (if someone knows how, they can post it) so we don't even have any circular logic. Even if there isn't, $\frac d{dx}\sin x=\cos x$ was proven sometime without referencing the L'Hopital's rule so we know it is true. Why wouldn't we then freely apply the L'Hopital's rule to $\frac {\sin x}x$?

PS I'm not saying that this is the best method to derive the limit or anything, but that I don't understand why it is so frowned upon and often considered invalid.

  • 7
    $\begingroup$ Once you've historically shown the limit / derivative without l'Hopital, you are principally allowed to use it here as well. Just don't do it before you ever have established what the derivative of $\sin x $ is. Then again, $\lim_{x\to0}\frac {\sin x}x=\cos 0 =1$ seems to use once limit rule less. $\endgroup$ Commented Jan 28, 2017 at 23:59
  • 4
    $\begingroup$ It's OK to do it, as long as you don't mind looking silly when you do. $\endgroup$
    – zhw.
    Commented Jan 29, 2017 at 0:04
  • $\begingroup$ Why would someone look silly when they do it? $\endgroup$
    – Blaza
    Commented Jan 29, 2017 at 0:22
  • $\begingroup$ It's like Tex Avery's kangaroo, hiding in its own pouch when it's frightened… $\endgroup$
    – Bernard
    Commented Jan 29, 2017 at 0:53
  • 2
    $\begingroup$ Because it's then clear the someone doesn't really know why that limit is $1.$ Instead, someone appeals to a result that depends on the limit being $1$ to show the limit is $1.$ Our someone is starting to sound like the butt of a Monty-Python joke. Why are such someones silly? Ask John Cleese. I had a litte rant on this topic here: math.stackexchange.com/questions/1286699/… $\endgroup$
    – zhw.
    Commented Jan 29, 2017 at 0:59

5 Answers 5


It is a correct application of l'Hôpital's rule, but using it to prove that the derivative of $\sin x$ is $\cos x$ is possibly circular logic, depending on your definition of $\sin x$.

If we did not know what the derivative of $\sin x$ was, but we did know the angle sum formula, here are some steps we could take to compute that derivative, starting from first principles, i.e. the definition of the derivative.

$$ (\sin x)'= \lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h} = \lim_{h\to 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\ = \cos x\cdot\left(\lim_{h\to 0}\dfrac{\sin h}{h}\right)+ \sin x\cdot\left(\lim_{h\to 0}\dfrac{\cos h -1}{h}\right) $$

To complete the computation of the derivative of $\sin x$, we must first know the limit of $\sin h/h$ and $(\cos h-1)/h$. We can't use l'Hôpital at this point because to use l'Hôpital you need to know the derivative of $\sin x$ at $0$, which is the very thing we are trying to compute. Assuming what you're trying to prove is a logical error known as "begging the question". Mathematics is axiomatic, so that every result builds on previous results, and no arguments are circular.

But if we can confirm the derivative of $\sin x$ by some other means, for example by defining it in terms of its Taylor series or a differential equation, then we may use it in a l'Hôpital computation to derive $\lim \sin x/x$ without fear of being circular. While not logically incorrect, this would still be fairly redundant: if you know the derivative of $\sin x$ just observe that, by definition, $\lim_{h\to 0}\sin x/x$ is just the derivative at 0. Using the machinery of l'Hôpital is overkill to get an answer you already know.

However if you're in a situation where you don't care about logical foundations and rigor, you're not proving theorems from first principles about calculus of trigonometric functions, you just need to compute $\lim\sin x/x$ and want to allow all known formulas and techniques, feel free to use l'Hôpital. It is correct.

  • $\begingroup$ If using l'Hopital to prove that the derivative of $\sin$ is $\cos$ is circular logic, how is using knowledge of the derivative of $\sin$ in order to use l'Hopital not circular? $\endgroup$
    – Arthur
    Commented Jan 29, 2017 at 0:14
  • $\begingroup$ It isn't because you don't need l'Hopital to prove that the derivative is $\cos x$ $\endgroup$
    – Blaza
    Commented Jan 29, 2017 at 0:17
  • $\begingroup$ @Blaza hi I added some details to the answer. Is it clearer why it is circular? You do need to know limit of $\sin h/h$, to compute derivative of $\sin x$, and using lhopital to compute it is circular. $\endgroup$
    – ziggurism
    Commented Jan 29, 2017 at 0:46
  • 6
    $\begingroup$ Supposing that you know the derivative of $\sin x$ is $\cos x$ by some means. Then it is a very roundabout way (but correct) to use L'Hospital's Rule to find the limit under question because by definition that limit is the derivative of $\sin x$ at $x=0$ and that is $\cos 0=1$. Remember that L'Hospital's Rule should not be used to calculate the limits of type $$\lim_{x\to a} \frac{f(x) - f(a)} {x - a} $$ because in the act of finding derivative of numerator you have already computed this particular limit and hence why go further. $\endgroup$
    – Paramanand Singh
    Commented Jan 29, 2017 at 4:04
  • $\begingroup$ @ParamanandSingh yes, that's a good point. $\endgroup$
    – ziggurism
    Commented Jan 29, 2017 at 4:23

I expand my comment to ziggurism's answer here.

Suppose that by a lot of hardwork, patience and ... (add more nice words if you like) I have obtained these three facts:

  1. $\cos x$ is continuous.
  2. $\dfrac{d}{dx}(\sin x) = \cos x$
  3. L'Hospital's Rule

and my next goal is establish the following fact: $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{*}$$ Because of all that hardwork, patience and ... I know that my goal $(*)$ is an immediate consequence of just the fact $(2)$ alone stated above. Then why would I combine all the three facts mentioned above to achieve my goal? To borrow an idea from user "zhw.", wouldn't a person doing this would be considered silly?

More often than not, many students don't really understand what's going behind the scenes when we use the mantra of "differentiate and plug" championed by L'Hospital's Rule. The act of differentiation itself entails that we know certain limits (and rules of differentiation) and further most of the derivatives are continuous (so that plugging works after differentiation step).

If one is so fond of L'Hospital's Rule why not put that to a better use to solve complex problems (like this and this) instead of using it to obtain limits which are immediate consequences of differentiation formulas.

  • $\begingroup$ Yeah, I know, it's like killing a fly with a canon. It's silly in that regard, but not necessarily incorrect, that was my point. Thanks for the answer. $\endgroup$
    – Blaza
    Commented Jan 29, 2017 at 10:01
  • 5
    $\begingroup$ @Blaza: the situation is not like using canon to kill a fly, but rather it's like having killed the fly (by squeezing it into your hands) and then using canon on the fly's corpse. $\endgroup$
    – Paramanand Singh
    Commented Jan 29, 2017 at 10:48
  • $\begingroup$ That's quite a picturesque analogy. I agree, thanks $\endgroup$
    – Blaza
    Commented Jan 29, 2017 at 11:11
  • $\begingroup$ @Blaza: actually I am not that an expert in creating analogies but I am happy that I was able to get my point across via some sort of analogy. And I am glad that you found it picturesque. $\endgroup$
    – Paramanand Singh
    Commented Jan 29, 2017 at 15:32
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    $\begingroup$ Nice post. +1. "Then why would I combine all the three facts mentioned above to achieve my goal?" In order to get ($*$) as an immediate consequence of (2) one needs to recognize that (i) $\frac{\sin x}{x}=\frac{\sin x- \sin 0}{x-0}$ (ii) by definition, $\sin'(0)=\lim_{x\to 0}\frac{\sin x- \sin 0}{x-0}$; inexperienced students may find applying L'Hopital more "straightforward" than identifying (i) and (ii). $\endgroup$
    – user587192
    Commented Dec 22, 2018 at 15:45

My personal favorite definition of sine and cosine come from (I think) Apostol's book, where he says

  1. $\sin, \cos : \Bbb R \to [-1, 1]$
  2. $\sin, \cos$ are continuous
  3. $\cos(\pi/2) = 0$
  4. $\cos(a-b) = \cos(a) \cos (b) + \sin(a) \sin(b)$

...and perhaps there's a fourth one. From these, it's easy to show (pick $a = b = 0$) that $$ \cos(0) = 1 \\ $$ and then picking $a = b$, to get that $$ \sin^2(t) + \cos^2(t) = 1 $$ for all $t$,

and picking $a = -b$, to derive the double-angle formula for cosine, and from that, the half-angle formula, and similar ones for sine.

From these, you can establish that if there are functions that satisfy these rules, then we can compute their values on $S = \Bbb Q\pi$, the set of all rational multiples of $\pi$. It's not too hard to do that, either. And then, with somewhat greater effort, you can show that these restrictions to $S$ are continuous.

Now you have continuous functions on $S$, which is evidently dense in $\Bbb R$, and hence there are unique continuous extensions of these functions to $\Bbb R$.

Finally, the cosine and sine addition and subtraction rules can be used to show that on $S$ (and hence on $\Bbb R$ as well), $\sin x \approx x$ for $x$ small, whence the limit we all love turns out to be one.

I'm pretty sure the details are all in Apostol's calculus.

An alternative approach is to define arcsine and arccos via integrals, observe that they're increasing with nonzero derivatives, and therefore have differentiable inverses which we define to be sine and cosine.

Any one of these avoids using L'h's rule to evaluate the limit. Hence using it to evaluate the limit later is just fine.

Of course, both these approaches require that you have some topology under your belt before you start calculus, and/or that you're willing to postpone the definition of trig functions until after you've defined integration, so they're not typical. That doesn't, however, make them wrong.


How do you prove that $\frac{d}{dx}[\sin(x)]=\cos(x)$? If you use the limit $$\lim_{x\to 0}\frac{\sin(x)}{x}=1 \qquad\qquad(1.1)$$ than you can't use the L'Hopital rule, because it is an example of circular logic.

You can avoid it, definying $$\sin(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\ \ \ \ \forall x\in\mathbb{R}$$ and $$\cos(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}\ \ \ \forall x\in\mathbb{R}$$.

Once you proved that $\frac{d}{dx}[\sin(x)]=\cos(x)$ and $\cos(0)=1$ than you can use the l'Hopital rule to calculate the limit (1.1).


Proving $\mathrm D \sin=\cos$ does not necessarily involve the use of l'Hopital if you define sine as a power series like Rudin.


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