trigonometric limit factoring

Question

I have this problem that I can't solve:

$$\lim_{x \to 0^+} \frac{3x-\sin 2x}{\sqrt{1-\cos x}}$$

Tried to multiply by the conjugate, so I can get rid of the squareroot on the denominator, but I am stuck after that. Can't use L'Hopital, only factoring:

$$\frac{3x-\sin 2x}{\sqrt{1-\cos x}}\cdot \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}=\frac{(3x-\sin 2x)\sqrt{1+\cos x}}{\sqrt{1-\cos^2 x}}$$

rewrite using trig identities and cancel the sqroot on the denominator. After that, I don't know how to continue:

$$\frac{(3x-\sin 2x)\sqrt{1+\cos x}}{\sqrt{\sin^2 x}}$$

PS: Any book that help with this types of problems? Thanks

Answer 1

Step by step :

$$ \begin{align*} \lim_{x\to0+}\frac{3x-\sin2x}{\sqrt{1-\cos x}} &=\lim_{x\to0+}\frac{(3x-\sin2x)(\sqrt{1+\cos x})} {(\sqrt{1-\cos x})(\sqrt{1+\cos x})}\\ &=\lim_{x\to0+}\frac{(3x-\sin2x)(\sqrt{1+\cos x})}{\sqrt{1-\cos^2x}}\\ &=\lim_{x\to0+}\frac{(3x-\sin2x)(\sqrt{1+\cos x})}{\sqrt{\sin^2x}}\\ &=\lim_{x\to0+}\frac{(3x-\sin2x)(\sqrt{1+\cos x})}{|\sin x|}\\ &=\lim_{x\to0+}\frac{(3x-\sin2x)(\sqrt{1+\cos x})}{\sin x}\\ &=\lim_{x\to0+}\frac{3x-\sin2x}{\sin x}(\sqrt{1+\cos x})\\ &=\lim_{x\to0+}\frac{3-\frac{\sin2x}x}{\frac{\sin x}x}(\sqrt{1+\cos x})\\ &=\frac{3-\lim\frac{\sin2x}x}{\lim\frac{\sin x}x}(\lim\sqrt{1+\cos x})\\ &=\frac{3-2}{1}\sqrt{1+1}=\sqrt2 \end{align*} $$

Note that $$\lim_{x\to0+}\frac{\sin x}x=1.$$

Answer 2

Continue from where you're stuck; since $x\to0^+$ you have: $$\lim_{x\to0^+}\frac{(3x-\sin 2x)\sqrt{1+\cos x}}{\sqrt{\sin^2 x}}=\lim_{x\to0^+}\frac{(3x-\sin 2x)\sqrt{1+\cos x}}{\sin x}$$ Now, I assume both limits exist and are finite: $$\lim_{x\to0^+}\frac{(3x-\sin 2x)\sqrt{1+\cos x}}{\sin x}=3\lim_{x\to0^+}\frac{x}{\sin x}\sqrt{1+\cos x}-\lim_{x\to 0^+}\frac{\sin 2x}{\sin x}\sqrt{1+\cos x} $$ The first limit is $3 \sqrt 2$, whereas the second is $2\sqrt 2$: $$\lim_{x\to 0^+}\frac{\sin 2x}{\sin x}\sqrt{1+\cos x}=2\lim_{x\to 0^+}\frac{\sin 2x}{2x}\frac{x}{\sin x}\lim_{x\to 0^+}\sqrt{1+\cos x}=2\sqrt 2 $$ Hence you conclude that: $$\lim_{x\to0^+}\frac{(3x-\sin 2x)\sqrt{1+\cos x}}{\sqrt{\sin^2 x}}=3\sqrt 2 -2\sqrt 2=\sqrt 2 $$