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# Evaluate the following limits in Exercises 1 to 22: limₓ→π/₂ tan 2x / [x - π/2]

**Solution:**

At x = π/2, the value of the given function takes the form 0/0, which is an indeterminate form.

So we will evaluate the given limit differently.

Now, put x - π/2 = y so that x → π/2, y → 0

Therefore,

limₓ→π/₂ tan 2x / [x - π/2]

= limᵧ→₀ tan 2(y + π/2)/y

= limᵧ→₀ tan (π + 2y)/y

= limᵧ→₀ tan 2y/y [∵ tan (π + 2y) = tan 2y]

= limᵧ→₀ sin2y/(y cos2y)

= limᵧ→₀ (sin2y/2y × 2/cos2y)

= (limᵧ→₀ sin2y/2y) × (limᵧ→₀ 2/cos2y) [y → 0 ⇒ 2y → 0]

= 1 x 2/cos 0 [∵ lim_{x→0} (sinx/x) = 1]

= 1 × 2/1

= 2

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1 Question 22

## Evaluate the following limits in Exercises 1 to 22: limₓ→π/₂ tan 2x / [x - π/2]

**Summary:**

The value of the limit limₓ→π/₂ tan 2x / [x - π/2] is 2

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