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$$\lim_{x\to 0} \frac{\sin(2x)}{4x}$$

In this form, it would be undefined, so how would you change it so that the denominator would not be $0$?

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    $\begingroup$ What are your thoughts on the problem? Have you discussed in class or read in your textbook about $\lim_{x \to 0} \frac{\sin x}{x}$? $\endgroup$ Sep 6, 2014 at 20:46
  • $\begingroup$ Do you know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$? Draw pictures. $\endgroup$ Sep 6, 2014 at 20:46
  • $\begingroup$ I did not know that $\endgroup$
    – user169562
    Sep 6, 2014 at 20:56

3 Answers 3

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As an alternative method, you could note that $\sin(2x)=2x+\mathcal{O}(x^3)$.

Then, $$ \lim_{x\to0}\frac{\sin(2x)}{4x}=\lim_{x\to0}\frac{2x+\mathcal{O}(x^3)}{4x}=\lim_{x\to0}\frac{1}{2}+\mathcal {O}(x^2)=\frac{1}{2}. $$

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$$\lim_{x \to 0} \frac{\sin (2x)}{4x}=\frac{1}{2} \lim_{x \to 0} \frac{\sin(2x)}{2x}=\frac{1}{2}$$

It is known that $\displaystyle\lim_{y \to 0} \frac{\sin y}{y}= 1$ (you can prove it using L'Hospital's Rule)

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    $\begingroup$ L'Hôpital's rule would not constitute a proof since we need this limit in order to prove that the derivative of $\sin$ is $\cos$. There is, however, a canonical proof for this using the squeeze theorem. $\endgroup$ Sep 6, 2014 at 20:48
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    $\begingroup$ Actually, $\displaystyle \lim_{y \rightarrow 0} \frac{\sin y}y = 1$. $\endgroup$
    – Cookie
    Sep 6, 2014 at 20:57
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    $\begingroup$ Oh,yes..I accidentally wrote that it is equal to $0$.. $\endgroup$
    – evinda
    Sep 6, 2014 at 20:58
  • $\begingroup$ @Omnom There are many other ways to define $\sin$ and $\cos$ such as the solution to a differential equation and as the inverse function of integrals. In each of these approaches we do not need the limit to prove that the derivative of $\sin$ is $\cos$. I believe the standard definition (as per most real analysis books) is in terms of the exponential function which means again that $\sin$ is readily differentiable. I agree that LH should not be used but it is generally false to call it circular. $\endgroup$
    – user157227
    Sep 6, 2014 at 21:15
  • $\begingroup$ @user157227 I had a similar argument here, which I'm not eager to repeat. I will simply state that, to the best of my knowledge, the trigonometric definition of $\sin(x)$ is the more common one to start with, certainly in the context of introductory calculus. $\endgroup$ Sep 6, 2014 at 21:22
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An unconventional solution would be with Maclaurin series;

The Maclaurin series of $$\frac {\sin(2x)}{4x}$$ is $$\frac 12-\frac{x^2}{3}+\frac{x^4}{15}+\mathcal{O}(x^6)$$ All the terms but the $\frac12$ become $0$ so there you go.

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