Limit of sin(3x)/sin(5x) as x→0

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Homework Statement


Find \lim_{x\rightarrow 0} {\frac{\sin(3x)}{\sin(5x)}}


The Attempt at a Solution


I know that the limit equals 0.6 (by typing it into my calculator), but I have no idea how to prove this, or even where to start. I know that sin is continuous, so I theoretically should be able to just plug it in, but obviously this doesn't work because it isn't divisible by 0.
 
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Try multiplying by 3x/(3x) and 5x/(5x), and placing the numerators and denominators strategically. The basic idea is that \lim_{u \to 0} \frac{sin u}{u} = 1
 
Thank you! I can't believe I didn't think of that answer - that helped me figure out the later questions as well.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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