Question

use lhôpitals rule to find the following limit.

\\( \lim _ { x \
ightarrow - \pi ^ { - } } ( x + \pi ) \cot x \\)

\\( \lim _ { x \
ightarrow - \pi ^ { - } } ( x + \pi ) \cot x = \\) (type an exact answer.)

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:

\(1\)