Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

$\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2} \frac{\pi}{3}-x}$


$\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2} \frac{\pi}{3}-x}$


Given $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}$

Now by using the formula

$\cos 6 x=1-2 \sin ^{2} 3 x$

Then the above equation becomes,

$\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-\left(1-2 \sin ^{2} 3 x\right)}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}$

Again using $\sin 3 x$ formula the above equation can be written as

$\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\left(1-2 \sin ^{2} 3 x\right)}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{2}|\sin 3 x|}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}$

Now we have

$\sin 3 x=\sin (\pi-3 x)=\sin 3\left(\frac{\pi}{3}-x\right)$

$\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}|\sin 3 x|}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin 3\left(\frac{\pi}{3}-x\right)}{\left(\frac{\pi}{3}-x\right)}$

On simplifying we get

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin 3\left(\frac{\pi}{3}-x\right)}{\left(\frac{\pi}{2}-x\right)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{3 \sin 3\left(\frac{\pi}{3}-x\right)}{3\left(\frac{\pi}{2}-x\right)}=3 \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin (\pi-3 x)}{(\pi-3 x)}$

Now as $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

By substituting the limit we get

$\Rightarrow$$3 \lim _{x \rightarrow \frac{t}{2}} \frac{\sin (\pi-3 x)}{(\pi-3 x)}=3.1=3$

$\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=3$

Leave a comment

Free Study Material