Calculus and Analysis
What the exam tests
Limits, continuity, differentiability, derivatives and their applications such as monotonicity, extrema and asymptotes, and integration including antiderivatives, definite integrals and areas. The exam favours a limit that needs algebra first and a function study that requires a sign table.
Core rules
- Factor, rationalise or use standard limits before L'Hôpital; L'Hôpital applies only to 0/0 or ∞/∞ forms.
- Standard limits: lim(x→0) sin x/x=1, lim(x→0)(1−cos x)/x²=½, lim(x→0)(e^x−1)/x=1, lim(x→0) ln(1+x)/x=1.
- Continuity at a requires three things: f(a) is defined, the limit exists, and the limit equals f(a).
- IVT: if f is continuous on [a,b] and f(a)f(b)<0, there is at least one root in (a,b).
- If f'(x)>0 on an interval, f is strictly increasing there; f'(x)=0 with a sign change gives an extremum.
- Integration: substitution reverses the chain rule; by parts uses ∫u dv=uv−∫v du; add +C for indefinite integrals.
Worked example
Find lim(x→0)(1−cos x)/x². Use 1−cos x=2 sin²(x/2). Then (1−cos x)/x² = 2 sin²(x/2)/x² = ½·[sin(x/2)/(x/2)]². As x→0, sin(x/2)/(x/2)→1, so the limit is ½. L'Hôpital twice gives the same value: sin x/(2x)→½, but the identity route shows why the limit is finite.
Traps
- Applying L'Hôpital to a limit that is not indeterminate, such as (x+1)/x at x=1.
- Differentiating |x| at 0 as if it were smooth.
- Omitting the inner derivative in a chain-rule calculation.
- Forgetting +C, or dropping the sign when reversing limits in a definite integral.
- Confusing a horizontal asymptote with a value the function actually reaches.