Answer:
L=24
Step-by-step explanation:
L'Hopital's Rule for Evaluating Limits:
Rule is that if [tex]\lim_{x \to a} \frac{f(x)}{g(x)}[/tex] takes [tex]\frac{0}{0}[/tex] or [tex]\frac{\infty}{\infty}[/tex] form, then,
[tex]\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}[/tex]
where [tex]f'(x)=\frac{df(x)}{dx}[/tex] and [tex]g'(x)=\frac{dg(x)}{dx}[/tex]
Now coming to the problem,
[tex]L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}[/tex]
Here [tex]f(x)=cot^{3}x-3cotx[/tex] and [tex]g(x)=cos(x+\frac{\pi}{3} )[/tex]
Substituting [tex]x=\frac{\pi}{6}[/tex] in f(x) and g(x),
[tex]f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0[/tex]
[tex]g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0[/tex]
Since L takes the form [tex]\frac{0}{0}[/tex], using l'hopital's rule
[tex]L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}[/tex]
now substituting [tex]x=\frac{\pi}{6}[/tex] ,
[tex]L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24[/tex]
