Find the value of $\lim \limits_{x \to \pi /3} \frac{(1-\cos6x)^{1/2}}{\sqrt 2 (\frac \pi3 - x)}$

Question

The question from my textbook requires us to find the value of the below limit:

$$\lim \limits_{x \to \pi /3} \frac{(1-\cos6x)^{1/2}}{\sqrt 2 (\pi/3 - x)}$$

On using the trigonometric identity $\cos2x = 1-2\sin^2x$, the above limit reduces to:

$$\lim \limits_{x \to \pi /3} \frac{\lvert \sin3x \rvert}{\pi/3 - x}$$

On substituting $\frac \pi3 - x$ by $y$ the above limit becomes:

$$\lim \limits_{y \to 0} \frac{\lvert \sin3y \rvert}y$$

In this case, the left hand and right hand limits are unequal, being equal to $-3$ and $3$ respectively. Thus, $$\lim \limits_{x \to \pi /3} \frac{(1-\cos6x)^{1/2}}{\sqrt 2 (\pi/3 - x)} \text{ does not exist.}$$ But my textbook gives the answer as $3$. It was apparently because it considered $(1-\cos6x)^{1/2}$ as equal to $\sin 3x$ and not as equal to $\lvert \sin3x \rvert$. But since $x \to \pi/3$, shouldn't $(\sin^2 3x)^{1/2}$ equal $\lvert \sin3x \rvert$? Is my argument incorrect? Any help would be appreciated.

Answer 1

Your analysis is correct; for example, see this graph: (the blue vertical line is at $\frac{\pi}{3}$)

Answer 2

HINT

Let $x=y+\pi/3$

$$\lim \limits_{x \to \pi /3} \frac{(1-\cos6x)^{1/2}}{\sqrt 2 (\frac \pi3 - x)}=\lim \limits_{y \to 0} \frac{(1-\cos6y)^{1/2}}{-\sqrt 2 y }$$

then use standard limit

• $\frac{1-\cos x}{x^2}\to \frac12$

and observe that

• $y>0 \implies y=\sqrt {y^2}$
• $y<0 \implies y=-\sqrt {y^2}$