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Isn't $\lim_{x \to 0} \frac{\sin 2x^{\circ}}{x^{\circ}}=2$? But the actual answer doesn't seem to have this rather answers of the form of radians.Where did i go wrong?

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    $\begingroup$ $2$ is the right answer. $\endgroup$ Jul 20, 2021 at 8:36
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    $\begingroup$ What is the source of the "answers" you refer to? Btw $2$ is correct. $\endgroup$ Jul 20, 2021 at 8:40
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    $\begingroup$ What is the superscript on $x$ for? Degrees? If it is degrees, then you will have to modify it to radians, which is where the conversionn comes in $\endgroup$
    – Alan
    Jul 20, 2021 at 8:40

2 Answers 2

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For an angle of $a$ radians and $x$ degrees it is $\dfrac{a}{\pi}=\dfrac{x}{180}\Rightarrow x=\dfrac{180a}{\pi} $ or $a=\dfrac{\pi x}{180}$

In your limit set $a=\dfrac{\pi x}{180}$. Since $x\to 0$ it will be $a\to 0$ and therefore

$\lim\limits_{x\to 0^{\circ}}\dfrac{sin(2x^{\circ})}{x^{\circ}}=\lim\limits_{a\to 0}\dfrac{sin(2\frac{180a}{\pi})}{\frac{180a}{\pi}}=\dfrac{2\frac{180}{\pi}}{\frac{180}{\pi}}=2$

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  • $\begingroup$ If I put into my calculator, while in degree mode, $\sin(0.02)/0.01$ it gives $0.034907$, or about $2\pi/180$. Your mistake is that the numerator of your limit is $\sin_{deg}(2x^{\circ})=\sin_{rad}(2a)$ so the limit is $\frac{2}{180/\pi}$. $\endgroup$ Jul 20, 2021 at 9:39
  • $\begingroup$ @Jaap Scherphuis I suppose that $sin_d(2x^{\circ})=sin_r(2a)$ is not actually true. However, there seems to be a difference between the limits $\lim\limits_{x\to0}\dfrac{\sin{(2x^{\circ})}}{x^{\circ}}$ and $\lim\limits_{x\to0}\dfrac{\sin{(2x^{\circ})}}{x}$. The second one is a real number whereas in the first one I suppouse that I have to add units, i.e. the result will be in $1/\text{degrees}$. Your calculator gives $\lim\limits_{x\to0}\dfrac{\sin{(2x^{\circ})}}{x}$ whereas I calculated $\lim\limits_{x\to0}\dfrac{\sin{(2x^{\circ})}}{x^{\circ}}$. The difference is a factor $\dfrac{180}{\pi}$ $\endgroup$
    – 1123581321
    Jul 20, 2021 at 11:03
  • $\begingroup$ My previous comment was not very clear. I'm saying that $\sin_d(2x^{\circ})=\sin_r(2a)$ is true, and that where you wrote $\sin(2\frac{180a}\pi)$ you are still using the $\sin_d$ function, but that you should be switching over to the $\sin_r$ function instead. So $\sin_d(2x^{\circ})=\sin_d(2\frac{180a}\pi)=\sin_r(2a)$. That is why you get an erroneous $\frac{180}\pi$ in the numerator of your limit. Try it with an explicit value for $x$ and $a$ and you'll see that $\sin(2x^{\circ})=\sin(2\frac{180a}\pi)$ is not changing the parameter value to radians, just rewriting it but still in degrees. $\endgroup$ Jul 20, 2021 at 11:16
  • $\begingroup$ @Jaap Scherphuis Ok maybe I misread your comment. My apologies. However it would be better if you could write your approach in detail as an answer. I'll try to think about the whole problem again. $\endgroup$
    – 1123581321
    Jul 20, 2021 at 12:27
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Since you can convert degrees to radians, by use of the formula $a_r = \frac{\pi}{180} \cdot a_d$, the expression becomes:

$\lim_{x \rightarrow 0} \frac{2 \sin(u) \cos(u)}{u}$, where $u = \frac{x\pi}{180}$, so clearly this is also equal to $\lim_{u \rightarrow 0} \frac{2\sin(u) \cos(u)}{u}$. Using the fact that $\lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 1$, the expression becomes $\lim_{u \rightarrow 0} 2\cos(u)$, which is clearly equal to $2$

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