Question

The function $f\left(x\right)=(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4})$ not defined at $x=0$
Let $L$ be the value of the function at $x=0$ so that it is continuous at $x=0$, then find the value of $L$

Answer 1

$f\left(x\right)=(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4})$

$\underset{x\to 0}{lim}f\left(x\right)={lim}_{x\to 0}(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4})$

$=\underset{x\to 0}{lim}\frac{(2+\cos x)x-3\sin x}{x^4\sin x}$

Applying L Hospital's Rule

$=\underset{x\to 0}{lim}\frac{2+\cos x-x\sin x-3\cos x}{{4x}^3\sin x+x^4\cos x}$

$=\underset{x\to 0}{lim}\frac{-\sin x-\sin x-x\cos x+3\sin x}{12x^2\sin x+4x^3\cos x+4x^3\cos x-x^4\sin x}$ (L-Hospita;ls' rule)

$=\underset{x\to 0}{lim}\frac{\sin x-x\cos x}{12x^2\sin x+8x^3\cos x-x^4\sin x}$

$=\underset{x\to 0}{lim}\frac{\cos x-\cos x+x\sin x}{24x\sin x+12x^2\cos x+24x^2\cos x-8x^3\cos x-x^4\cos x-4x^3\sin x}$

$=\underset{x\to 0}{lim}\frac{1}{24+\frac{36x}{\tan x}-\frac{8x^2}{\tan x}-\frac{x^3}{\tan x}-4x^2}$

Dividing both numerator and denominator by $x\sin x$

$=\frac{1}{24+36-\frac{8.1}{1}-0-0}$

$=\frac{1}{60}$