e. $\\lim_{x\\to 0}\\left(\\frac{1}{\\sin x}-\\frac{1}{x}\\right)$
e. $\\lim_{x\\to 0}\\left(\\frac{1}{\\sin x}-\\frac{1}{x}\\right)$
Answer
Explanation:
Step1: Combine the fractions
$$ \begin{align*} \lim_{x\rightarrow0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)&=\lim_{x\rightarrow0}\frac{x - \sin x}{x\sin x}\ \end{align*} $$
Step2: Apply L'Hopital's Rule (since it is in $\frac{0}{0}$ form)
Differentiate numerator and denominator: $$ \begin{align*} \lim_{x\rightarrow0}\frac{x - \sin x}{x\sin x}&=\lim_{x\rightarrow0}\frac{1-\cos x}{\sin x + x\cos x}\ \end{align*} $$ Still in $\frac{0}{0}$ form, apply L'Hopital's Rule again.
Step3: Differentiate numerator and denominator again
$$ \begin{align*} \lim_{x\rightarrow0}\frac{1-\cos x}{\sin x + x\cos x}&=\lim_{x\rightarrow0}\frac{\sin x}{\cos x+\cos x - x\sin x}\ \end{align*} $$
Step4: Substitute $x = 0$
$$ \begin{align*} \lim_{x\rightarrow0}\frac{\sin x}{2\cos x - x\sin x}&=\frac{\sin0}{2\cos0-0\times\sin0}\ &=\frac{0}{2 - 0}\ &=0 \end{align*} $$
Answer:
$0$