Limits, Continuity and Differentiation
This domain tests evaluation of limits, the continuity test at a point, and differentiation by the standard rules including chain, product, quotient and implicit forms.
Limits
Standard results: lim(x→0) sin x / x = 1, lim(x→0) (aˣ − 1)/x = log a, lim(x→0) (1+x)^(1/x) = e, lim(x→0) (1−cos x)/x² = 1/2. For 0/0 forms, factorise, rationalise, or use the standard limits.
Continuity and differentiability
f is continuous at x = a when LHL = RHL = f(a). Differentiability at a implies continuity there, but continuity does not imply differentiability (|x| at 0 is the standard counterexample).
Derivatives to know
- (xⁿ)′ = nx^(n−1), (sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = sec²x, (eˣ)′ = eˣ, (log x)′ = 1/x.
- Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v². Chain: d/dx f(g(x)) = f′(g(x))·g′(x).
Worked example
lim(x→0) (1 − cos x)/x². Use 1 − cos x = 2 sin²(x/2): the expression becomes 2 sin²(x/2)/x² = (1/2)·[sin(x/2)/(x/2)]². As x→0 the bracket tends to 1, so the limit is 1/2. For a derivative, y = sin(x²) gives dy/dx = cos(x²)·2x = 2x cos(x²) by the chain rule.
Traps that cost marks
- Differentiating the outer function but forgetting the inner derivative.
- Sign errors on (cos x)′ and (cot x)′.
- Reversing the quotient rule numerator to uv′ − u′v.
- Concluding "not continuous" from "not differentiable", or assuming continuity gives differentiability.
- Computing a limit by direct substitution when the form is 0/0.