# $\lim_{x\to 0}\frac{\sin(20x)}{\sin(301x)}$

$\lim_{x\to 0}\frac{\sin(20x)}{\sin(301x)}$

A very simple one. Intuitively I know the answer must be $\frac{20}{301}$, but a don't have the slightest idea of how to manipulate this function algebraically in order to get rid of the $\frac{0}{0}$ as $x$ goes to $0$. A hint would be awesome. But I'm seeking for a solution without the use of tools such as L'Hospital.

Thank you very much!

• – Simply Beautiful Art Apr 1 '17 at 11:55
• I don't get our intuition. Did you cancel out sines ? – A---B Apr 1 '17 at 11:57
• Of course not. It was just because the numerator and the denominator are both going to 0, but the denomitator is doing so 301/20 times faster. But of course this "intuition" is nothing but intuition without a formal proof. – R. Maia Apr 1 '17 at 13:45

We know that $\lim_{x\to 0}\frac{\sin x}x=1$, thus we can simply multiply and divide by $20x$ and $301x$, in this way: $$\lim_{x\to 0}\frac{\sin(20x)}{\sin(301x)}=$$ $$=\lim_{x\to 0}\frac{\sin(20x)}{\sin(301x)}\cdot\frac{20x}{20x}\cdot\frac{301x}{301x}=$$ $$=\lim_{x\to 0}\frac{\sin(20x)}{20x}\cdot\frac{20x}{301x}\cdot\frac{301x}{\sin(301x)}=$$ $$=\lim_{x\to 0}\frac {20x}{301x}=\frac{20}{301}$$

Hint :

$$\lim_{x\to 0}\frac{\sin(20x)}{\sin(301x)}=\lim_{x\to 0} (\frac{\sin 20x}{20x}\times \frac{301x}{\sin 301 x}\times \frac{20x}{301x} )$$

Hint:

$$\frac{\sin(ax)}{\sin(bx)}=\frac{\frac{\sin(ax)}{ax}}{\frac{\sin(bx)}{bx}}\frac ab$$

$\lim_{x\to 0}\frac{\sin(20x)}{\sin(301x)} \cdot \frac{301 \cdot 20 \cdot x}{301 \cdot 20 \cdot x}$

$=\lim_{x\to 0}\frac{\sin(20x)}{20 \cdot x} \cdot \frac{301 \cdot x}{\sin(301x)} \cdot \frac{20}{301}$

$=\lim_{x\to 0}\frac{\sin(20x)}{20 \cdot x} \cdot \frac{1}{\frac{\sin(301x)}{301 \cdot x}} \cdot \frac{20}{301}$

$=1 \cdot 1 \cdot \frac{20}{301}$

$=\frac{20}{301}$

• Any doubt feel free to ask. – Kanwaljit Singh Apr 1 '17 at 11:53