Question

use lhôpitals rule to find the following limit
lim (2 sec x - 2 tan x)
x→π/2^-

Explanation:

Step1: Rewrite the expression

First, rewrite \(2\sec x - 2\tan x\) using trigonometric identities. Since \(\sec x=\frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\), we have \(2\sec x - 2\tan x=\frac{2 - 2\sin x}{\cos x}\). As \(x
ightarrow\frac{\pi}{2}^{-}\), this is in the \(\frac{0}{0}\) indeterminate form.

Step2: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator. The derivative of \(2 - 2\sin x\) is \(- 2\cos x\), and the derivative of \(\cos x\) is \(-\sin x\). So, \(\lim_{x
ightarrow\frac{\pi}{2}^{-}}\frac{2 - 2\sin x}{\cos x}=\lim_{x
ightarrow\frac{\pi}{2}^{-}}\frac{-2\cos x}{-\sin x}\)

Step3: Evaluate the limit

Substitute \(x = \frac{\pi}{2}\) into \(\frac{-2\cos x}{-\sin x}\). We know that \(\cos\frac{\pi}{2}=0\) and \(\sin\frac{\pi}{2} = 1\). So, \(\lim_{x
ightarrow\frac{\pi}{2}^{-}}\frac{-2\cos x}{-\sin x}=\frac{-2\times0}{-1}=0\)

Answer:

\(0\)