Question

Calculate the limit of the function with L'Hospital rule​

Answer 1

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ takes $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then,

$\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}$

where $f'(x)=\frac{df(x)}{dx}$ and $g'(x)=\frac{dg(x)}{dx}$

Now coming to the problem,

$L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}$

Here $f(x)=cot^{3}x-3cotx$ and $g(x)=cos(x+\frac{\pi}{3} )$

Substituting $x=\frac{\pi}{6}$ in f(x) and g(x),

$f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0$

$g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0$

Since L takes the form $\frac{0}{0}$, using l'hopital's rule

$L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}$

now substituting $x=\frac{\pi}{6}$ ,

$L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24$