IBDP Maths AHL 5.13 The evaluation of limits AA HL Paper 1- Exam Style Questions- New Syllabus

Method 1: L’Hôpital’s Rule
Direct substitution gives the indeterminate form \( \frac{0}{0} \). Applying L’Hôpital’s rule:
\( \lim_{x \to 0} \frac{x \sin x}{1 – \cos x} = \lim_{x \to 0} \frac{\frac{d}{dx}(x \sin x)}{\frac{d}{dx}(1 – \cos x)} = \lim_{x \to 0} \frac{\sin x + x \cos x}{\sin x} \).
This is still \( \frac{0}{0} \), so we apply the rule a second time:
\( = \lim_{x \to 0} \frac{\cos x + \cos x – x \sin x}{\cos x} = \lim_{x \to 0} \frac{2\cos x – x \sin x}{\cos x} \).
Substituting \( x = 0 \):
\( \frac{2(1) – 0(0)}{1} = 2 \).

Method 2: Trigonometric Identities
Multiply the numerator and denominator by the conjugate \( (1 + \cos x) \):
\( \lim_{x \to 0} \frac{x \sin x (1 + \cos x)}{(1 – \cos x)(1 + \cos x)} = \lim_{x \to 0} \frac{x \sin x (1 + \cos x)}{1 – \cos^2 x} \).
Using \( 1 – \cos^2 x = \sin^2 x \):
\( \lim_{x \to 0} \frac{x \sin x (1 + \cos x)}{\sin^2 x} = \lim_{x \to 0} \frac{x (1 + \cos x)}{\sin x} \).
Recognizing the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\( \lim_{x \to 0} \frac{1 + \cos x}{\left(\frac{\sin x}{x}\right)} = \frac{1 + 1}{1} = 2 \).

Method 3: Maclaurin Series Expansion
Using the expansions \( \sin x = x – \frac{x^3}{6} + \dots \) and \( \cos x = 1 – \frac{x^2}{2} + \frac{x^4}{24} – \dots \):
Numerator: \( x \sin x = x^2 – \frac{x^4}{6} + \dots \)
Denominator: \( 1 – \cos x = \frac{x^2}{2} – \frac{x^4}{24} + \dots \)
The ratio becomes:
\( \frac{x^2 – \frac{x^4}{6} + \dots}{\frac{x^2}{2} – \frac{x^4}{24} + \dots} = \frac{1 – \frac{x^2}{6} + \dots}{\frac{1}{2} – \frac{x^2}{24} + \dots} \).
As \( x \to 0 \), the terms with higher powers of \( x \) vanish:
\( \frac{1}{1/2} = 2 \).

Method 4: Half-Angle Identities
Using \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \) and \( 1 – \cos x = 2 \sin^2 \frac{x}{2} \):
\( \frac{x \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} = \frac{x \cos \frac{x}{2}}{\sin \frac{x}{2}} \).
Rearranging to use the standard limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \):
\( = 2 \cdot \left( \frac{x/2}{\sin(x/2)} \right) \cdot \cos(x/2) \).
As \( x \to 0 \), the limit is \( 2 \cdot (1) \cdot (1) = 2 \).

Final Answer:
\( \boxed{2} \)