Further Calculus — IB Math AA (HL)

1. L'Hôpital's Rule

L'Hôpital's rule evaluates indeterminate limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by differentiating numerator and denominator separately.

L'Hôpital's Rule

If $\displaystyle\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = 0$ (or $\pm\infty$), and $g'(x) \ne 0$ near $a$, then:

$$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$$

The rule may be applied repeatedly if the result remains indeterminate.

Worked Example 1

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$ using L'Hôpital's rule

Check form: At $x=0$: $\sin 0 = 0$ and $x=0$, so this is $\frac{0}{0}$ — indeterminate. Apply L'Hôpital.

Differentiate: $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\cos x}{1} = \frac{\cos 0}{1} = 1$

Result: $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = 1$

2. Maclaurin Series

The Maclaurin series is a Taylor series expanded about $x=0$. It expresses a function as an infinite polynomial.

Standard Maclaurin Series (First 4 Terms)

$e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$

$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$

$\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$

$\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots$ ($|x| \le 1$, $x\ne -1$)

$(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots$

Worked Example 2

Find the Maclaurin series for $e^{2x}$ up to the term in $x^3$

Substitute $2x$ for $x$ in $e^x$ series: $e^{2x} = 1 + (2x) + \dfrac{(2x)^2}{2!} + \dfrac{(2x)^3}{3!} + \cdots$

Simplify: $e^{2x} = 1 + 2x + \dfrac{4x^2}{2} + \dfrac{8x^3}{6} + \cdots = 1 + 2x + 2x^2 + \dfrac{4}{3}x^3 + \cdots$

3. Differential Equations

A differential equation (DE) relates a function to its derivatives. The general solution contains an arbitrary constant $C$; a particular solution uses an initial condition to fix $C$.

Separable Differential Equations

A DE of the form $\dfrac{dy}{dx} = f(x)g(y)$ is separable: rearrange to $\dfrac{1}{g(y)}dy = f(x)dx$ and integrate both sides.

Worked Example 3

Solve $\dfrac{dy}{dx} = xy$ with initial condition $y(0) = 2$

Separate variables: $\dfrac{1}{y}\,dy = x\,dx$ (assuming $y \ne 0$)

Integrate both sides: $\displaystyle\int \frac{1}{y}\,dy = \int x\,dx \Rightarrow \ln|y| = \frac{x^2}{2} + C_1$

General solution: $|y| = e^{C_1}e^{x^2/2}$, so $y = Ae^{x^2/2}$ where $A = \pm e^{C_1}$.

Particular solution: $y(0)=2 \Rightarrow A = 2$. Therefore $y = 2e^{x^2/2}$.

Integrating Factor Method (Linear First-Order DEs)

For $\dfrac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is $\mu(x) = e^{\int P(x)\,dx}$. Multiply both sides by $\mu$; the LHS becomes $\dfrac{d}{dx}[\mu y]$, then integrate.

Integration by Parts

$$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$$

LIATE rule for choosing $u$: Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential (choose $u$ from the earlier category).

Practice Problems

Q1. Evaluate $\displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}$ using L'Hôpital's rule.

Show Solution

$\frac{0}{0}$ form. Apply once: $\displaystyle\lim_{x\to 0}\frac{e^x-1}{2x}$ — still $\frac{0}{0}$. Apply again: $\displaystyle\lim_{x\to 0}\frac{e^x}{2} = \frac{1}{2}$.

Q2. Find $\displaystyle\int x e^x\,dx$ using integration by parts.

Show Solution

Let $u = x$, $dv = e^x dx$. Then $du = dx$, $v = e^x$.

$\displaystyle\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C$

Q3. Solve the DE $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$ for $x > 0$.

Show Solution

Integrating factor: $\mu = e^{\int 1/x\,dx} = e^{\ln x} = x$.

Multiply: $\dfrac{d}{dx}(xy) = x^3$. Integrate: $xy = \dfrac{x^4}{4} + C$. So $y = \dfrac{x^3}{4} + \dfrac{C}{x}$.

4. Exam Tips

L'Hôpital's rule can only be applied to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms — verify the form first.
For Maclaurin series, substitution (e.g., replacing $x$ with $2x$) is faster than computing derivatives from scratch.
In separable DEs, always check whether $y=0$ is also a solution (it may be excluded during division).
With integration by parts, if the integral reappears, collect terms on one side to solve algebraically (common with $e^x\sin x$ type).