Functions, Limits, Continuity and Differentiability

Calculus studies how functions change. Machine learning uses it everywhere: training a model means finding the minimum of a loss function, and that needs derivatives. This lesson builds the foundation: limits, continuity and differentiability.

Functions

A function f assigns to each x in its domain exactly one value f(x). Common facts GATE uses:

• Domain restrictions: √g(x) needs g(x) ≥ 0; ln g(x) needs g(x) > 0; 1/g(x) needs g(x) ≠ 0.
• Even functions satisfy f(−x) = f(x) (x², cos x, |x|); odd functions satisfy f(−x) = −f(x) (x³, sin x).
• A function is increasing on an interval if x₁ < x₂ ⇒ f(x₁) ≤ f(x₂), and strictly increasing with <.
• Composition: (f ∘ g)(x) = f(g(x)).

Limits

lim_x→a f(x) = L means f(x) gets arbitrarily close to L as x gets close to a (without being equal to a). The value f(a) itself, or whether it exists, does not matter.

• One-sided limits: lim_x→a⁻ (from the left) and lim_x→a⁺ (from the right). The two-sided limit exists if and only if both one-sided limits exist and are equal.
• Limits respect sums, products, quotients (when the denominator's limit is not 0) and continuous functions.

Standard limits (worth memorising):

Example. lim_x→2 (x² − 4)/(x − 2): at x = 2 this is 0/0, but for x ≠ 2 the fraction equals x + 2, so the limit is 4.

Limits at infinity of rational functions. Divide by the highest power of x in the denominator. If the degrees are equal, the limit is the ratio of the leading coefficients: lim_x→∞ (3x² + 1)/(5x² − x) = 3/5.

Indeterminate forms and L'Hôpital's rule

The forms 0/0, ∞/∞, 0·∞, ∞ − ∞, 0⁰, 1^∞ and ∞⁰ are indeterminate: the limit could be anything and must be worked out.

L'Hôpital's rule. If lim f(x)/g(x) has the form 0/0 or ∞/∞, then

lim f(x)/g(x) = lim f′(x)/g′(x), provided the right-hand limit exists.

Apply it repeatedly if needed. Other forms are rewritten first:

• 0 · ∞: write f · g as f/(1/g).
• 1^∞, 0⁰, ∞⁰: take logarithms. If y = f^g, then ln y = g ln f; find that limit L, and the answer is e^L.

Examples.

• lim_x→0 (eˣ − 1 − x)/x² = lim (eˣ − 1)/(2x) = lim eˣ/2 = 1/2.
• lim_x→0⁺ x ln x = lim (ln x)/(1/x) = lim (1/x)/(−1/x²) = lim (−x) = 0.
• lim_x→∞ (1 + 2/x)ˣ: ln y = x ln(1 + 2/x) → 2, so the limit is e².

Continuity

f is continuous at a if all three hold:

1. f(a) is defined,
2. lim_x→a f(x) exists,
3. lim_x→a f(x) = f(a).

Polynomials, eˣ, sin x and cos x are continuous everywhere; rational functions, ln x and √x are continuous on their domains. Sums, products, quotients (where defined) and compositions of continuous functions are continuous.

Types of discontinuity:

• Removable: the limit exists but differs from f(a) (or f(a) is undefined). Redefining f(a) fixes it.
• Jump: the one-sided limits exist but differ (a step function).
• Infinite: the function blows up, like 1/x at 0.
• Oscillating: no limit at all, like sin(1/x) at 0.

Intermediate value theorem. If f is continuous on [a, b] and k lies between f(a) and f(b), then f(c) = k for some c in [a, b]. In particular, if f(a) and f(b) have opposite signs, f has a root in between: this is why the bisection method works.

Extreme value theorem. A continuous function on a closed, bounded interval [a, b] attains a maximum and a minimum there.

Differentiability

The derivative of f at a is

f′(a) = lim_h→0 (f(a + h) − f(a)) / h

if the limit exists. It is the slope of the tangent line, or the instantaneous rate of change.

• Differentiable ⇒ continuous, but not the other way round.
• The classic counterexample is f(x) = |x| at x = 0: it is continuous, but the left derivative is −1 and the right derivative is +1. A corner (or a cusp, or a vertical tangent) means no derivative.
• At a boundary between two formulas, check that the function values match (continuity) and that the left and right derivatives match.

Example. For which a and b is f(x) = x² for x ≤ 1, and ax + b for x > 1, differentiable at 1? Continuity: 1 = a + b. Matching slopes: 2x at 1 is 2, so a = 2. Then b = −1.

import math

def limit(f, a, side=0, steps=(1e-2, 1e-4, 1e-6)):
    return [round(f(a + side * h if side else a + h), 6) for h in steps]

print("sin x / x near 0:", limit(lambda x: math.sin(x) / x, 0))
print("(1 - cos x)/x^2 near 0:", limit(lambda x: (1 - math.cos(x)) / x ** 2, 0, steps=(1e-1, 1e-2, 1e-3)))
print("(e^x - 1 - x)/x^2 near 0:", limit(lambda x: (math.exp(x) - 1 - x) / x ** 2, 0, steps=(1e-1, 1e-2, 1e-3)))
print("x ln x as x -> 0+:", limit(lambda x: x * math.log(x), 0, side=1))
print("(1 + 2/x)^x for large x:", [round((1 + 2 / x) ** x, 4) for x in (10, 1000, 100000)], " e^2 =", round(math.e ** 2, 4))

# |x| at 0: one-sided difference quotients disagree
for h in (0.1, 0.001):
    print(f"h = {h}: right quotient {(abs(h) - 0) / h:+.1f}, left quotient {(abs(-h) - 0) / -h:+.1f}")

sin x / x near 0: [0.999983, 1.0, 1.0]
(1 - cos x)/x^2 near 0: [0.499583, 0.499996, 0.5]
(e^x - 1 - x)/x^2 near 0: [0.517092, 0.501671, 0.500167]
x ln x as x -> 0+: [-0.046052, -0.000921, -1.4e-05]
(1 + 2/x)^x for large x: [6.1917, 7.3743, 7.3889]  e^2 = 7.3891
h = 0.1: right quotient +1.0, left quotient -1.0
h = 0.001: right quotient +1.0, left quotient -1.0

Practice questions