# Evaluate the following limits in Exercises 1 to 22: limₓ→π sin (π - x)/[π (π - x)]

**Solution:**

By substituting x =π, we get 0/0, which is an indeterminate form.

So we will evaluate the given limit differently.

It can be seen that x → π ⇒ (π - x) → 0

Therefore,

limₓ→π sin (π - x)/[π (π - x)]

= 1/π lim₍π ₋ ₓ₎→₀ sin (π - x)/ (π - x)

= (1/π) × 1 [∵ lim_{y→0} (siny/y) = 1]

= 1/π

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1 Question 15

## Evaluate the following limits in Exercises 1 to 22: limₓ→π sin (π - x)/[π (π - x)]

**Summary:**

The value of the limit limₓ→π sin (π - x)/[π (π - x)] is 1/π

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