# Evaluate the following limits in Exercises 1 to 22: limₓ→₀ (cos 2x -1)/(cos x - 1)

**Solution:**

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

So we will evaluate the given limit differently.

The given limit is

limₓ→₀ (cos2x - 1)/(cosx - 1)

Using the double angle formulas,

= limₓ→₀ (1 - 2 sin²x - 1)/(1 - 2 sin²(x/2) - 1) [∵ cosx = 1 - 2 sin²x/2]

= limₓ→₀ (sin²x)/(sin²x/2)

Multiplying and dividing the numerator by x^{2} and the denominator by (x/2)^{2,}

= limₓ→₀ [x² (sin²x/x²)] / [(x/2)^{2 }(sin²(x/2)/(x/2)^{2}]

= 4 [limₓ→₀ [(sin²x/x²)]/[lim₍ₓ/₂₎_{→0 }(sin²x/2)/(x/2)²]

= 4 [limₓ→₀ (sinx/x)]² / [lim₍ₓ/₂₎_{→0 }(sin x/2)/(x/2)²]² [x → 0 ⇒ x/2 → 0]

= 4 (1²/1²) [∵ lim_{y→0} (siny/y) = 1])

= 4

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1 Question 17

## Evaluate the following limits in Exercises 1 to 22: limₓ→₀ (cos 2x -1)/(cos x - 1)

**Summary:**

The value of the limit limₓ→₀ (cos 2x -1)/(cos x - 1) is 4