# Limits of trigonometric functions when x tends to 0

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?

• $2$ is the right answer. Jul 20, 2021 at 8:36
• What is the source of the "answers" you refer to? Btw $2$ is correct. Jul 20, 2021 at 8:40
• 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
– Alan
Jul 20, 2021 at 8:40

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

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

• 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 $\frac{2}{180/\pi}$. Jul 20, 2021 at 9:39
• @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 $\dfrac{180}{\pi}$ Jul 20, 2021 at 11:03
• 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. Jul 20, 2021 at 11:16
• @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. Jul 20, 2021 at 12:27

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 $$2$$