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Proof of lim (x to 0) of sinx/x and circular proofs

  1. Nov 21, 2009 #1
    I wish to prove [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] using L'Hôpital's rule. The problem with this is, even though the result after applying the rule is 1 (the correct answer), the limit itself was assumed to be correct in order to calculate the derivative of sinx. This constitutes circular reasoning, and thus is not a valid proof.

    In order to make this proof valid, we need to find a way to determine the derivative of sinx that does not rely on [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] to be true.

    If we use Euler's formula, we find that [itex]\sin x = \frac{e^{ix} - e^{-ix}}{2i}[/itex] and [itex]\cos x = \frac{e^{ix} + e^{-ix}}{2}[/itex].

    This means that [itex]\frac{d}{dx} \sin x = \frac{d}{dx} \frac{e^{ix} - e^{-ix}}{2i}[/itex]. But [itex]\frac{d}{dx} \frac{e^{ix} - e^{-ix}}{2i} = \frac{e^{ix}+e^{-ix}}{2}[/itex] (using the quotient rule) which is cosx. Therefore, [itex]\frac{d}{dx}\sin x = \cos x[/itex]. This was found without the assumption of [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] at all. This means we can now use L'Hôpital's rule to find the limit without constituting circular reasoning.

    Is my reasoning above correct? This proof would be invalid is Euler's formula relies on [itex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/itex] but I am not sure.
  2. jcsd
  3. Nov 21, 2009 #2
    Luckily, it is not necessary to evaluate this limit in order to find the derivative of sin(x) at x = 0. It is only necessary that sine is continuous at 0, that the derivative of sine exists in a neighborhood of x=0 but not necessarily at 0, and that the limit as x approaches 0 of the derivative function of sine also exists. A theorem in basic calculus states that the derivative at 0 is then equal to this limit (the derivative function is continuous there).
    Through other means, we find that the derivative function of sine when x is not 0 is cos(x), whose aforementioned limit as x approaches 0 is 1, completing the proof. The proof you presented above is perfectly valid.

    Euler's formula also relies on knowledge of the derivative of sine or defining sine by a differential equation. Either proof is valid.
  4. Nov 21, 2009 #3


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    Homework Helper

    This is a common question. You do not need L'Hôpital's rule as
    [itex]\lim_{x \to 0} \frac{\sin x}{x} = \sin'(0)=\cos(0)=1[/itex]
    by the definition of the derivative.
    You have not stated which definition of sine you are using. Several of then allow sin'(0) to be found without finding your limit first.
    Common definitions include
    My favorite definition
    4)lim_{x->0} sin(x)/x=1
    in which the limit is a given
    power series
    [tex]\sin(x):=\sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}[/tex]
    differential equation
  5. Nov 21, 2009 #4


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    Since they are the same thing I do not see how you can find one with out finding the other. Perhaps you mean there are several indirect ways to evaluate this limit in order to find derivative of sin(x) at x = 0.
  6. Nov 21, 2009 #5
    I state my meaning explicitly in the rest of the paragraph. The proof does not require the limit.
  7. Nov 21, 2009 #6
    Yes, this is true. Isn't Euler's formula derived using the Taylor expansion of sinx (which relies on the derivative of sinx)?
  8. Nov 21, 2009 #7
    That is one method of deriving Euler's formula. There are others, but they each require knowledge of the derivative of the sine function.
  9. Nov 21, 2009 #8
    Consider the following derivation of sin'(x):

    [itex]\frac{d}{dx}\sin x = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h}[/itex]
    [itex]= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}[/itex]
    [itex]= \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}[/itex]
    [itex]= \lim_{h \to 0} \frac{\sin x (\cos h - 1)}{h} + \lim_{h \to 0} \frac{\cos x \sin h}{h}[/itex]
    [itex] = \sin x (0) + \cos x (1)[/itex]
    [itex] = \cos x[/itex]

    See how the derivative of sinx relies on the fact that [itex]\lim_{x \to 0}\frac{\sin x}{x} = 1[/itex]. This is why I can't use L'Hôpital's rule to prove this limit, because this would involve taking the derivative of sinx which (as shown above) assumes what I am trying to prove!
  10. Nov 21, 2009 #9
    That derivation is not necessary. You can still use the theorem stated in my post.
    Specifically, if f is continuous at a, f'(x) exists in a neighborhood of a, and [itex]\lim_{x\rightarrow a} f'(x)[/itex] exists, then f'(a) exists and
    [tex]f'(a) = \lim_{x\rightarrow a} f'(x)[/tex]
    The fact that sin'(x) = cos(x) when x is not an integral multiple of pi requires a definition of sin(x): either a definition using arclength, area, or differential equations. In the derivation you have given, it is not apparent what either sin(x) or cos(x) have as a definition and thus you have to resort to algebra and the limit of sin(x)/x.
    Last edited: Nov 21, 2009
  11. Nov 21, 2009 #10
    What other means? I am only aware of the difference quotient (or whatever its called) to find the derivative of a function.
  12. Nov 21, 2009 #11
    There are many ways of defining sine. lurflurf gave several ways. One method is to define cos(t) to be the x-coordinate of the point on a unit circle whose radius makes an angle t with the positive x-axis, and then let sine be defined by [itex]\sin t = \sqrt{1 - \cos^2 (t)}[/itex] when t is between 0 and pi. This appeals to having sin(x) be the y-coordinate of the point in question on the Cartesian plane using Euclidean geometry. The periodic continuation is trivial. The important part is finding a symbolic expression for cos(t), which we can get by means of an integral, whose derivative is then easy to find (fundamental theorem of calculus). By the definition, sine is continuous everywhere, and thus the theorem above completes its derivative for the isolated points where x is a multiple of pi. This is one approach.
    Other approaches include a functional approach where the relationship between sine and cosine is defined by the sum and difference formulas for angles on a finite domain and continued periodically as above. This is given by lurflurf's formulas above.
    Other approaches still declare the existence of functions satisfying certain differential equations everywhere (thus assuming your derivative beforehand, making this approach boring for your proof), and then showing uniqueness of the solution.
    Whichever method you use, you then apply the derivative/calculus to that definition, which is usually more enlightening than trying to wrangle a black box called "sin(x)".
  13. Nov 21, 2009 #12
    You're right, I haven't given a definition of sin(x). Without knowing what exactly sin(x) is, the only information that can be used to find its derivative is the limit of sin(x)/x.

    My motivation for starting this thread was actually finding a purely algebraic (ie. non geometric) method of proving the limit of sin(x)/x. I didn't know that the derivative of sin(x) at x = 0 could be determined without knowing the limit of sin(x)/x. I guess it depends on what I define sin(x) to be, and the approach I use.

    If I know for sure that sin'(0) = 1 and sin(0) = 0 (whatever "sin(x)" is), then all I need to do is use the definition of the derivative:
    [itex]\lim_{h \to 0} \frac{\sin (0 + h) - \sin(0)}{h} = 1[/itex]
    ie [itex]\lim_{h \to 0} \frac{\sin h}{h} = 1[/itex]
    (This was stated in previous posts).
    Last edited: Nov 21, 2009
  14. Nov 21, 2009 #13
    Thankyou everyone for your help. :)
  15. Nov 22, 2009 #14
    I'm not sure the proof I'm giving here is relevant to the topic, but since there are tons of definition of the trigonometric functions with different proofs to find their properties ( which end up being the same at the end ), you might want to have a look at this one. This one might help if you have defined trigonometric with the help of the trigonometric circle.


    ( click the img, S1, S2, S3 are the areas in grey, x is the angle. The squeeze theorem is applied when the limits appear )
  16. Nov 22, 2009 #15


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    It is perfectly valid to define sin(x) to be
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!)}x^{2n+ 1}[/tex]
    and define cos(x) to be
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}[/tex]
    and, since those are power series with infinite radius of convergence, the derivatives follow from term-by-term differentiation.

    Equivalently, you can define sin(x) to be the solution to the initial value problem:
    y"= -y with y(0)= 0 and cos(x) to be the solution to the inital value problem y"= -y with y(0)= 1. Since y"= -y is a second order, linear, homogeneous equation, any solution can be written as a linear combination of two independent solutions. It is easy to see that sin(x) and cos(x) are independent and, in fact, if y(x) is any solution to y"= -y with y(0)= A, y'(0)= B, then y(x)= A cos(x)+ B sin(x).

    Since sin(x) satisfies a second order differential equation, it is twice differentiable. In particular, it is differentiable. Let y(x) be its derivative: y= (sin x)'. Since sin x is twice differentiable, y is differentiable and y'= (sin x)"= -sin(x) (because sin(x) satisfies y"= -y) and since sin(x) is differentiable, y is twice differentiable and y"= - (sin(x))'= -y. Furthermore, y(0)= (sin(x))' at 0 which is 1 and y'(0)= - sin(0)= 0. Therefore, y(x)= (1)cos(x)+ (0)sin(x)= cos(x)- without using that limit.
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