lim⁡x→π/2[xtan⁡x−(π2)sec⁡x]\lim_{x \rightarrow \pi/2}\left[x \tan x - \left(\frac{\pi}{2}\right) \sec x\right] is e…

Mathematics

Problem

lim⁡x→π/2[xtan⁡x−(π2)sec⁡x]\lim_{x \rightarrow \pi/2}\left[x \tan x - \left(\frac{\pi}{2}\right) \sec x\right] is equal to (a) 1 (b) -1 (c) 0 (d) none of these
Intuitive explanation

This limit involves an indeterminate form where both terms blow up as x→π/2x \to \pi/2. The key is to rewrite everything in terms of sin⁡x\sin x and cos⁡x\cos x, then use a Taylor expansion around x=π/2x = \pi/2 by substituting u=x−π/2u = x - \pi/2 (so u→0u \to 0). This transforms the messy trigonometric behavior into a polynomial approximation you can evaluate cleanly.

Intuitive explanation

This limit involves an indeterminate form where both terms blow up as x→π/2x \to \pi/2. The key is to rewrite everything in terms of sin⁡x\sin x and cos⁡x\cos x, then use a Taylor expansion around x=π/2x = \pi/2 by substituting u=x−π/2u = x - \pi/2 (so u→0u \to 0). This transforms the messy trigonometric behavior into a polynomial approximation you can evaluate cleanly.

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