Calculus

A limit describes what $f(x)$ approaches as $x$ gets arbitrarily close to a value $a$ — it says nothing about $f(a)$ itself, which is why limits can exist even where the function is undefined. Continuity at $x=a$ needs three things simultaneously: $f(a)$ is defined, the limit exists (both one-sided limits agree), and the limit equals $f(a)$. Vertical asymptotes appear when the denominator hits zero but the numerator doesn't; horizontal asymptotes come from limits as $x \to \pm\infty$, found by comparing the degrees of the top and bottom.

When direct substitution gives $\frac{0}{0}$, the limit can still exist — this is the single most tested limit technique at SL. Factorise (polynomials) or rationalise (surds), cancel the common factor, then substitute — write out the algebra explicitly, because 'show that' questions are pure method marks. The squeeze theorem rescues limits with a bounded oscillating factor multiplied by something shrinking to zero: if $g(x) \le f(x) \le h(x)$ near $x=a$ and both outer limits equal $L$, then $\lim_{x\to a} f(x)=L$ too.

The derivative $f'(x)$ is a gradient function built from the limit $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$, though you'll rarely use first principles unless a question explicitly demands it. Beyond the power rule, most real questions need the chain rule (derivative of the outside, inside left alone, times derivative of the inside), the product rule, or the quotient rule — and identifying the correct rule before substituting is half the battle. $f'(x)>0$ means increasing, $f'(x)<0$ means decreasing, and $f'(x)=0$ flags a stationary point that still needs classifying.

Where $f'(x)=0$, you have a stationary point — the second derivative classifies it: $f''(x)>0$ means concave up (local minimum), $f''(x)<0$ means concave down (local maximum). A point of inflection is where concavity actually changes sign either side, not just where $f''(x)=0$ — that condition alone is not enough. The second derivative test is faster than checking the sign of $f'$ on either side, but it's inconclusive whenever $f''(x)=0$, in which case you must fall back on the first derivative test.

Indefinite integration reverses differentiation: $\int f'(x)\,dx = f(x)+C$, where $C$ represents the whole family of curves sharing that gradient function — always substitute a given boundary point in AFTER integrating to pin down $C$. A definite integral $\int_a^b f(x)\,dx$ is a single number equal to $F(b)-F(a)$, representing signed area between the curve and the x-axis. If a question wants actual physical area (not signed), you must split at the roots of $f(x)$ and add the absolute value of each piece — the same logic gives total distance travelled as $\int_a^b |v(t)|\,dt$, split at every root of $v(t)$.