Q.33 The value of
lim_x→0 [(x − sin 2x) / (x − sin 5x)] (rounded off to two decimal places) is _________.

Limit Evaluation Steps

Direct substitution at x = 0 gives:

0 − 0 / 0 − 0 = 0/0

This is an indeterminate form.

First L’Hôpital application:

(1 − 2cos 2x) / (1 − 5cos 5x)

This is still 0/0 because cos 2x and cos 5x → 1 as x→0.

Second L’Hôpital gives:

(4 sin 2x) / (25 sin 5x)

Now apply standard limits:

lim x→0 (sin kx)/(kx) = 1

Rewriting:

(4/25) × (sin 2x)/(2x) × (2x)/(sin 5x/(5x)) × (5x/x) × (x/x) = 8/125

Final Answer:

Exact value = 8/125 = 0.064 ≈ 0.06

Taylor Series Method

Expand sine terms:

sin 2x

2x − (8x³/6) + O(x⁵)

So

x − sin 2x = −x + (4x³/3) + O(x⁵)

sin 5x

5x − (125x³/6) + O(x⁵)

So

x − sin 5x = −4x + (125x³/6) + O(x⁵)

Taking the ratio of cubic leading terms:

(4/3)x³ / (125/6)x³ = 8/125 = 0.064 ≈ 0.06

Common Mistakes Explained

Mistake 1: Treating sin2x/sin5x ≈ 2/5 = 0.40 (ignores missing x terms)

Mistake 2: Stopping after one L’Hôpital gives: (1−2)/(1−5) = 0.20 (incomplete)

Mistake 3: Using sin(kx) ≈ kx gives (x−2x)/(x−5x) = 1/4 = 0.25 (misses x³ accuracy)

Summary

The correct limit is:

8/125 = 0.064 ≈ 0.06