\(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\) equals
- (a)\(0\)
- (b)\(2\)
- (c)does not exist
- (d)\(4\)
Revision notes, 17 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
Calculus unit: 8 of 80 theory marks (Limits and Derivatives).
\(\displaystyle\lim_{x \to a} f(x) = L\) means \(f(x)\) gets as close as we like to \(L\) as \(x\) gets close to \(a\) (\(x \neq a\)). The limit exists when the left-hand limit equals the right-hand limit. Limits of sums, differences, products and quotients (non-zero denominator) are the sums, differences, products and quotients of the limits.
\(\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x} = 0\) (\(x\) in radians). Hence \(\displaystyle\lim_{x \to 0} \frac{\sin ax}{x} = a\) and \(\displaystyle\lim_{x \to 0} \frac{\tan x}{x} = 1\).
\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]
It is the rate of change of \(f\), and the slope of the tangent to \(y = f(x)\). Finding it from this definition is "from first principles".
Find \(\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\).
\(\dfrac{(x - 3)(x + 3)}{x - 3} = x + 3 \to 6\).
Find \(\displaystyle\lim_{x \to 0} \frac{\sin 4x}{\sin 2x}\).
\(\dfrac{\sin 4x}{4x} \cdot \dfrac{2x}{\sin 2x} \cdot 2 \to 2\).
Differentiate \(y = x^3\cos x\).
\(y' = 3x^2\cos x - x^3\sin x\).
Topics in this chapter: Polynomial and rational functions · Rules and standard derivatives · Trigonometric limits · The derivative · Limits.
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
You can evaluate standard limits by substitution or factorising and differentiate polynomials and sin, cos.
You can use sin x/x → 1, differentiate from first principles and apply the product and quotient rules.
You can handle limits needing surds or trigonometric rewriting and use derivatives as rates of change.
You can find constants that make a limit exist and prove standard derivatives from the definition.
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 2 of the 17 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
\(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\) equals
If \(f(x) = 3x^2 - 5x + 2\), then \(f'(1)\) is
\(\displaystyle\lim_{x \to 0} \frac{\sin 3x}{x}\) equals
Want it against the clock? Take a timed 30-mark chapter test on Limits and Derivatives, new questions each time (CBSE Essentials or the free trial).