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$$\lim_{h\to 0}\frac{\sin(h)}{h}$$

I wanna ask whether the value of this limit equal to 1 or not if the "h" is used in degrees and not in radians ? My teacher told me that it's not equal to 1 if "h" is used in degree but I wanna confirm because he was also not so sure.

Dávid Laczkó
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Ankit
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    do you mean $\displaystyle \lim_{h \to 0}\frac{\sin h}{h} $? – user2628206 Aug 03 '21 at 18:28
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    Your teacher's right. See the answers here: https://math.stackexchange.com/questions/1339540/why-does-the-derivative-of-sine-only-work-for-radians – Hans Lundmark Aug 03 '21 at 18:31
  • What is $h°$ supposed to do in the denominator then? – Aman Kushwaha Aug 03 '21 at 18:32
  • Then you can consider $\frac{\sin(2\pi (d/360))}{2\pi (d/360)}$ with $d$ in degrees. – Michael Aug 03 '21 at 18:32
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    Note that $\displaystyle \lim_{h \to 0}\frac{\sin h^\circ}{h}$ i not of the same form as $\displaystyle \lim_{h \to 0}\frac{\sin h^\circ}{h^\circ}$, the later of which is equivalent to $\displaystyle \lim_{h \to 0}\frac{\sin h}{h}$. – user2628206 Aug 03 '21 at 18:37
  • @user2628206 so $$lim_{h\rightarrow 0}\frac{sin(h°)}{h°}=1$$ ?? – Ankit Aug 03 '21 at 18:43
  • @Ankit Yes, if using degrees as formulated in Sangchul Lee 's response. Also I have a typo in previous comment: i should be is! – user2628206 Aug 03 '21 at 18:52

2 Answers2

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In mathematics, each trigonometric function only takes pure numbers as arguments, and the argument is geometrically interpreted as radian.1) Even the unit $\text{rad}$ is defined in SI units purely as the number $1$.

So, what is the degree then? You can simply consider the degree symbol ${}^{\circ}$ as the constant having the value

$${}^{\circ}=\frac{\pi}{180},$$

i.e., it is simply a conversion factor. Then

  1. If you regard $\sin$ to take degree as the input, then it is the same as considering $\sin(x^{\circ})$, where ${}^{\circ}$ is the above constant. In this case,

    $$ \lim_{h\to 0} \frac{\sin(h^{\circ})}{h} = {}^{\circ} = \frac{\pi}{180}. $$

  2. If you simply replace $h$ by $h$ degree, i.e. $h^{\circ}$, then nothing changes because

    $$ \lim_{h\to 0} \frac{\sin(h^{\circ})}{h^{\circ}} = 1. $$

1) This interpretation is valid only when the argument is real. The trigonometric functions can take complex arguments as well, and then the geometric interpretation no longer applies.

Sangchul Lee
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    But now you have $h$ playing two roles. In the numerator, you've replaced $h$ by $ah$, so now that $h$ is $180/\pi$ times bigger than the $h$ in the denominator. – B. Goddard Aug 03 '21 at 18:43
  • @B.Goddard, I think I might have misinterpreted OP's claim. So I now presented two versions of answers depending on the interpretation. Thank you for pointing this out. – Sangchul Lee Aug 03 '21 at 18:45
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    He says $h$ is measured in degrees. So isn't the bottom $h$ also counting how many degrees? I think this is more of a problem of being sloppy with what $\sin x$ means. – B. Goddard Aug 03 '21 at 18:49
  • @Sangchul Lee thanks for your second edit I was asking for the second part you added where h is in degree in both num. and deno.. – Ankit Aug 03 '21 at 18:51
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I'll write $\operatorname{Sin}h^\circ$ with a capital $\text{S}$ and let $\sin,$ with a lower-case $\text{s}$ mean sine in radians. Then $$ \operatorname{Sin}h^\circ = \sin\frac{\pi h}{180}. $$ So: $$ \frac{\operatorname{Sin}h^\circ} h = \frac{\sin(\pi h/180)} h = \frac\pi{180}\cdot\frac{\sin(\pi h/180)}{\pi h/180} = \frac\pi{180}\cdot\frac{\sin u} u. $$

Michael Hardy
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