Limit of sin function

Question

I need some help solving the following limit:

Here is how I started:

$$\lim_{x \to 0^+}\frac{\sin(6x)}{\sqrt{\sin(2x)}} = \lim_{x \to 0^+}\frac{\sin(6x)\sqrt{\sin(2x)}}{{\sin(2x)}}=\lim_{x \to 0^+}\frac{\frac{6 \sin(6x)}{6x}\frac{\sqrt{\sin(2x)}}{x}}{{\frac{2 \sin(2x)}{2x}}}$$

I don't know how to deal with this sqrt...

Thanks! :)

Answer 1

Your last step has an error. The $x$ under the square root shouldn't be there. So your next step should be

$$= \lim_{x\to 0^+} \frac{6}{2}\sqrt{\sin 2x} = 0.$$

Answer 2

The limit equals:

$$\lim_{x \to 0^+} \frac{\sin(6x)}{6x} \frac{1}{\sqrt{\frac{\sin(2x)}{2x}}} \cdot 6x \cdot \frac{1}{\sqrt{2x}} = \lim_{x \to 0^+} 1 \cdot 1 \cdot 6x \cdot \frac{1}{\sqrt{2x}} = \lim_{x \to 0^+} \frac{6}{\sqrt{2}} \sqrt{x} = 0.$$

Answer 3

Substitute $y=2x\to 0^+$:

$$\lim_{y\to 0^+} \frac{\sin 3y}{\sqrt{\sin y}}=\lim_{y\to 0^+} \frac{3\sin y-4\sin^3 y}{\sqrt{\sin y}} =\lim_{y\to 0^+} 3\sqrt{\sin y}-4(\sin y)^{5/2} =0 $$

Answer 4

Use the asymptotic equivalence $\sin(x) \sim x$ for $ c\rightarrow 0^{+}$.

Hence you get $$\lim_{x \rightarrow 0^{+}} \frac{6x}{\sqrt{2x}} = \lim_{x \rightarrow 0^{+}} \frac{6\sqrt{x}}{\sqrt{2}} =0 $$