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Without using L'Hospital's Rule, why does $$\lim_{x\to 0} \frac{\sin x}{x} = 1?$$

Em.
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1 Answers1

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Let's use the Taylor Series expansion of $\sin x$ and see if we can get a limit. The reason we do this is because when we divide by $x$, we are essentially dividing by a polynomial function. For convenience, let's convert $\sin x$ to a polynomial function too !

Here's a trick question I like a lot, not entirely related to the question here. Is any transcendental equation (Involving $\log$, trigonometric, exponential functions) linear ? The answer is no ! Because they all have a series expansion in higher powers of $x$ !

$$\begin{align} \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + {\dots}\\ \implies \frac{\sin x}{x} &= 1 -\frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + {\dots}\\ \implies \lim_{x\to 0} \frac{\sin x}{x} &= 1 -\frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + {\dots}\\ \implies \lim_{x\to 0} \frac{\sin x}{x} &= 1 - 0 \\ \implies \lim_{x\to 0} \frac{\sin x}{x} &= 1 \\ \end{align}$$

Saikat
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    Any argument that uses the derivative of $sin$ is circular because this limit is necessary to prove that the derivative of $sin$ is $cos$. – John Douma Feb 27 '16 at 04:41
  • @JohnDouma So, which is axiomatic in your opinion the limit or the derivative ? – Saikat Feb 27 '16 at 05:31
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    Neither. The limit can be proved using a simple geometric argument along with the squeeze theorem. The limit can then be used to calculate the derivatives of the trig functions which can then be used with Taylor's theorem to expand the sine function into its Taylor series. – John Douma Feb 27 '16 at 06:00
  • @JohnDouma The $\sin$ function may be defined as a power series. This answer is perfectly valid. – MathematicsStudent1122 Feb 27 '16 at 06:16
  • @MathematicsStudent1122 No it can not. The sine function is a trig function so if you claim a series is equal to the sine function you must prove that. – John Douma Feb 27 '16 at 06:40
  • @JohnDouma When you say the sine function is a trig function, you are simply asserting that the trigonometric definition is the one you're working with. There is nothing stopping us from working with a series definition especially when angles and triangles are largely irrelevant in analysis. Remember, you can use a variety of definitions to introduce various functions. $\exp$, for example, is often introduced as a power series, and using that fact one may show that it is its own derivative without any circularity. – MathematicsStudent1122 Feb 27 '16 at 06:53
  • @MathematicsStudent1122 So then if I define $sin{x}$ to be $0$ for all $x$ the limit is equal to $0$. – John Douma Feb 27 '16 at 06:58
  • @JohnDouma To be sine, it needs to satisfy certain properties that we want the sine function to have. These can be derived from a power series definition, but not your "definition". In particular, we can show that it is equivalent to other definitions. – MathematicsStudent1122 Feb 27 '16 at 07:03
  • I agree with @JohnDouma ...people sometimes make the assertion that this is a valid definition of sine but it only technically is. I have never seen it introduced this way and am quite sure I have never met anyone who was introduced to $sin(x)$ in this way, and I would bet a lot of money the OP wasn't either, making this solution perfectly invalid. – JLA Feb 27 '16 at 07:11
  • @JLA Statements like "it only technically is"..."I have never seen"... "I would bet a lot of money" are not mathematically relevant. If you want to make inferences about the what definition the OP is working with, that is fine, but it is something else to say that the answer is "circular" and thus wrong. – MathematicsStudent1122 Feb 27 '16 at 07:15
  • @MathematicsStudent1122 Do you doubt that the definition the OP had intended isn't the standard one (the geometric one)? Like almost any question there is room for interpretation, and there is only one obvious way to interpret this question, I believe. The OP wants their question answered, not someone's misinterpretation of the question. If I'm right, then even if this answer is "right", it's not useful, so what's the point? – JLA Feb 27 '16 at 07:17
  • Gosh ! OP asked for a proof without L'Hopital's rule. He didn't say without calculus or derivatives. Yes, the Taylor Series needs derivatives and limits for its definition. But, those limits may be computed in methods other than L'Hopitals's rule. – Saikat Feb 27 '16 at 08:06
  • I don't think Mathematics needs utility to justify itself. The 'point' is in the elegance of how we can use an infinite series expansion to describe a ratio in triangles and how we can choose different points of view about the same thing to solve different problems ! – Saikat Feb 27 '16 at 08:12