TheIBAxolotl
Continuity: no lifting the pen
Informally, a function is continuous at \(x = a\) if you can draw its graph through that point without lifting your pen. More precisely, three things must hold:
- \(f(a)\) exists,
- \(\lim_{x \to a} f(x)\) exists (the values approach the same number from both sides),
- the two are equal.
Typical failures: a hole (the limit exists but \(f(a)\) is missing or different), a jump (left and right limits differ), or a vertical asymptote.
Limits that converge or diverge
A limit converges if it approaches a finite value and diverges otherwise. Examples: \(\lim_{x \to \infty} \frac{1}{x} = 0\) converges; \(\lim_{x \to 0^+} \frac{1}{x}\) grows without bound, so it diverges. One-sided limits (\(x \to a^-\) and \(x \to a^+\)) are useful for piecewise functions.
Differentiability: no corners
A function is differentiable at \(a\) if the tangent gradient exists there, meaning the chord gradients approach the same value from both sides. Differentiability needs continuity, but continuity is not enough: \(f(x) = |x|\) is continuous at 0 but has a sharp corner there (gradient \(-1\) on the left, \(+1\) on the right), so it is not differentiable at 0.
Worked example: piecewise function
\(f(x) = x^2\) for \(x \le 1\) and \(f(x) = ax + b\) for \(x > 1\). Find \(a\) and \(b\) so that \(f\) is differentiable everywhere.
- Continuity at 1: left value \(1^2 = 1\) must equal right value \(a + b\), so \(a + b = 1\).
- Matching gradients at 1: left derivative \(2x = 2\), right derivative \(a\), so \(a = 2\).
- Hence \(b = -1\), and the line \(y = 2x - 1\) is exactly the tangent to \(x^2\) at \(x = 1\).
First principles
The formal definition of the derivative is
\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.\]
It is the chord gradient with the second point a distance \(h\) away, with \(h\) shrunk to zero.
Example. For \(f(x) = x^3\):
\[\frac{(x + h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2 \to 3x^2.\]
Example. For \(f(x) = \frac{1}{x}\):
\[\frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \frac{x - (x + h)}{h\,x(x + h)} = \frac{-1}{x(x + h)} \to -\frac{1}{x^2}.\]
The essential step is always to simplify until the \(h\) in the denominator cancels before letting \(h \to 0\).
Higher derivatives
Differentiating again gives the second derivative \(f''(x)\) or \(\frac{d^2y}{dx^2}\), and so on: \(f^{(n)}(x)\) or \(\frac{d^n y}{dx^n}\). For \(y = x^4\), \(y' = 4x^3\), \(y'' = 12x^2\), \(y''' = 24x\), \(y^{(4)} = 24\), and every higher derivative is 0.
Common mistakes
- Substituting \(h = 0\) before cancelling, which gives \(\frac{0}{0}\).
- Forgetting the limit notation in each line of a first-principles proof.
- Assuming continuity implies differentiability.
- Checking only one condition (value or gradient) in piecewise questions.
Exam tips
- Expand \(f(x + h)\) fully and show the cancellation of \(f(x)\) terms.
- Write "\(\lim_{h \to 0}\)" until the final line; marks are lost if it disappears early.
- In piecewise questions, write both conditions as equations, then solve.