evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim x→0− 3(1/x - csc…

Explanation:

Step1: Rewrite the expression

First, rewrite $\csc x=\frac{1}{\sin x}$. So the limit becomes $\lim_{x\rightarrow0^{-}}3\left(\frac{1}{x}-\frac{1}{\sin x}\right)=\lim_{x\rightarrow0^{-}}\frac{3(\sin x - x)}{x\sin x}$. As $x\rightarrow0^{-}$, we have the $\frac{0}{0}$ - form, so L'Hopital's Rule can be applied.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $3(\sin x - x)$ is $3(\cos x - 1)$ and the derivative of $x\sin x$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v=\sin x$ is $\sin x+x\cos x$. So the limit is $\lim_{x\rightarrow0^{-}}\frac{3(\cos x - 1)}{\sin x+x\cos x}$. This is still in the $\frac{0}{0}$ - form, so apply L'Hopital's Rule again.

Step3: Apply L'Hopital's Rule again

Differentiate the numerator and denominator again. The derivative of $3(\cos x - 1)$ is $- 3\sin x$, and the derivative of $\sin x+x\cos x$ is $\cos x+\cos x - x\sin x=2\cos x - x\sin x$. So the limit is $\lim_{x\rightarrow0^{-}}\frac{-3\sin x}{2\cos x - x\sin x}$.

Step4: Evaluate the limit

Substitute $x = 0$ into $\frac{-3\sin x}{2\cos x - x\sin x}$. We know that $\sin(0)=0$ and $\cos(0)=1$. So $\lim_{x\rightarrow0^{-}}\frac{-3\sin x}{2\cos x - x\sin x}=\frac{-3\times0}{2\times1-0\times0}=0$.

Answer:

$0$