use lhôpitals rule to find the following limit.\n\n\\( \\lim _ { x \\rightarrow - \\pi ^ { - } } ( x + \\pi…

Explanation:

Step1: Rewrite the function

We know that (\cot x=\frac{\cos x}{\sin x}), so ((x + \pi)\cot x=\frac{(x + \pi)\cos x}{\sin x}). When (x\to-\pi^{-}), we have (\lim_{x\to-\pi^{-}}(x + \pi)=0) and (\lim_{x\to-\pi^{-}}\sin x=0). So, this is in the (\frac{0}{0}) indeterminate form, and we can apply L'Hôpital's Rule.

Step2: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator. The derivative of the numerator (u=(x + \pi)\cos x) using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u=x+\pi), (u^\prime = 1), (v=\cos x), (v^\prime=-\sin x). So (u^\prime=(x + \pi)\cos x) derivative is (\cos x-(x + \pi)\sin x). The derivative of the denominator (v = \sin x) is (\cos x). By L'Hôpital's Rule, (\lim_{x\to-\pi^{-}}\frac{(x + \pi)\cos x}{\sin x}=\lim_{x\to-\pi^{-}}\frac{\cos x-(x + \pi)\sin x}{\cos x}).

Step3: Evaluate the limit

Substitute (x =-\pi) into (\frac{\cos x-(x + \pi)\sin x}{\cos x}). When (x=-\pi), (\cos(-\pi)=- 1), ((x+\pi)\sin x = 0). So (\lim_{x\to-\pi^{-}}\frac{\cos x-(x + \pi)\sin x}{\cos x}=\frac{\cos(-\pi)-0}{\cos(-\pi)} = 1).

Answer:

(1)