use lhôpitals rule to find the following limit\nlim (2 sec x - 2 tan x)\nx→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\rightarrow\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\rightarrow\frac{\pi}{2}^{-}}\frac{2 - 2\sin x}{\cos x}=\lim_{x\rightarrow\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\rightarrow\frac{\pi}{2}^{-}}\frac{-2\cos x}{-\sin x}=\frac{-2\times0}{-1}=0)

Answer:

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