Calculus: Limits, Continuity and Differentiation
What this domain tests
Limits of rational, irrational, exponential, logarithmic and trigonometric expressions; continuity at a point and on an interval; the intermediate value theorem; derivatives as limits of difference quotients; tangent lines, monotonicity, extrema, concavity and optimisation, plus Rolle's theorem and the mean value theorem.
Limits and asymptotes
- Substitute first. If you get 0/0, factorise, multiply by the conjugate, or use standard limits: lim(x→0) sin x / x = 1, lim(x→0)(e^x − 1)/x = 1, lim(x→0) ln(1+x)/x = 1, lim(x→0)(1+x)^(1/x) = e.
- At infinity divide by the highest power of x; a ratio of polynomials behaves like the ratio of leading terms.
- Asymptotes: x = a if f → ±∞ there; y = L if lim f = L; oblique y = ax + b with a = lim f(x)/x and b = lim (f(x) − ax).
Continuity and differentiation
- f is continuous at a when lim(x→a) f(x) = f(a); both one-sided limits must exist and agree.
- IVT: if f is continuous on [a, b] and f(a), f(b) have opposite signs, some c has f(c) = 0.
- f′(a) = lim(h→0) (f(a+h) − f(a))/h. Differentiable implies continuous, not conversely: |x| at 0 is the standard counterexample.
- Product, quotient and chain rules; (ln x)′ = 1/x, (e^x)′ = e^x, (sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = 1/cos²x.
- Rolle: continuous on [a, b], differentiable on (a, b), f(a) = f(b) ⟹ some c has f′(c) = 0. MVT: f′(c) = (f(b) − f(a))/(b − a).
Worked example
Compute lim(x→0) (1 − cos 3x)/x². Substitution gives 0/0. Use 1 − cos θ = 2 sin²(θ/2) with θ = 3x, so the numerator is 2 sin²(1.5x). Then
(1 − cos 3x)/x² = 2 · [sin(1.5x)/x]² = 2 · [1.5 · sin(1.5x)/(1.5x)]².
As x → 0 the bracket tends to 1.5 · 1 = 1.5, whose square is 2.25, so the limit is 2 · 2.25 = 4.5 = 9/2.
Traps that cost marks
- Replacing sin(3x) by 3x when sin(3x) is a term of a sum, not a factor.
- Quoting IVT or Rolle without verifying continuity (and differentiability on the open interval).
- Differentiating a product as the product of the derivatives, or dropping the inner factor: (sin 3x)′ = 3cos 3x.
- Treating every solution of f′(x) = 0 as an extremum, without checking that f′ changes sign.
- Losing one-sided behaviour at the join of a piecewise function.