Question

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.
\\(lim_{x
ightarrowpi}(x - pi)cot x\\)
\\(lim_{x
ightarrowpi}(x - pi)cot x=square\\) (type an exact answer.)

Explanation:

Step1: Rewrite the function

First, rewrite \((x - \pi)\cot x\) as \(\frac{x-\pi}{\tan x}\). As \(x
ightarrow\pi\), we have the indeterminate - form \(\frac{0}{0}\), so L'Hopital's Rule can be applied.

Step2: Differentiate the numerator and denominator

Differentiate the numerator \(u=x - \pi\), \(u^\prime=1\), and the denominator \(v = \tan x\), \(v^\prime=\sec^{2}x\) according to the derivative rules.

Step3: Find the limit of the new - function

By L'Hopital's Rule, \(\lim_{x
ightarrow\pi}\frac{x - \pi}{\tan x}=\lim_{x
ightarrow\pi}\frac{1}{\sec^{2}x}\).

Step4: Evaluate the limit

Since \(\sec x=\frac{1}{\cos x}\), when \(x
ightarrow\pi\), \(\cos\pi=- 1\), then \(\sec\pi=-1\) and \(\sec^{2}x = 1\). So \(\lim_{x
ightarrow\pi}\frac{1}{\sec^{2}x}=1\).

Answer:

1