Q.$\lim_{x \to 0} \dfrac{1 - \cos 4\theta}{1 - \cos 6\theta}$ is
(A) $\dfrac{4}{9}$
(B) $\dfrac{1}{2}$
(C) $\dfrac{-1}{2}$
(D) $-1$

A Toolkit of Named Limits

Some limits recur so often across problems that it is worth memorising their values outright, along with the one theorem that proves the trickiest of them: the Sandwich (Squeeze) Theorem.

The Sandwich Theorem

Theorem 9.5. If $g(x) \le f(x) \le h(x)$ for all $x$ near $x_0$ (except possibly at $x_0$ itself), and if

then $\lim_{x\to x_0} f(x) = l$ too — $f$ is "squeezed" between two functions that agree in the limit, so it has no room to do anything else.

Illustration: to show $\displaystyle\lim_{x\to 0} x^2\sin\!\frac{1}{x^2} = 0$, note that $\sin(\cdot)$ is always between $-1$ and $1$, so $-x^2 \le x^2\sin\frac{1}{x^2} \le x^2$. Since both $-x^2$ and $x^2$ tend to $0$ as $x\to 0$, the Sandwich Theorem forces the middle expression to $0$ as well — even though $\lim_{x\to 0}\sin\frac{1}{x^2}$ on its own does not exist (it oscillates wildly), so the product rule alone could never have been applied directly.

The two flagship trigonometric limits

Result 9.1.

Part (a) is proved geometrically by sandwiching $\frac{\sin\theta}{\theta}$ between $\cos\theta$ and $1$ using the areas of a triangle, a sector, and a larger triangle built on the unit circle; since both bounding functions tend to $1$ as $\theta\to0$, so must $\frac{\sin\theta}{\theta}$. Part (b) follows algebraically from (a) by writing $1-\cos\theta = 2\sin^2\frac{\theta}{2}$ and splitting the quotient into a $\sin$-over-argument piece (which uses part (a)) times a factor that vanishes.

A direct corollary worth keeping separate: $\displaystyle\lim_{x\to 0}\sin x = 0$, obtained from the sandwich $-|x|\le \sin x \le |x|$.

The full standard-limit toolkit (§9.2.10)

Alongside the trig pair above, these are worth having on instant recall — none require anything beyond algebra and substitution to use (their proofs, where given, lean on the exponential/log relationship or on the trig pair):

And the three equivalent forms of the number $e$ as a limit:

The recognise-and-substitute pattern

Nearly every "hard-looking" limit in this section is really one of the above standard forms in disguise, reached via a clean substitution $y = (\text{some expression in } x)$ chosen so that $y \to 0$ (or $y\to\infty$) exactly when $x$ does. The book's worked examples all follow this shape:

1. Spot the shell. Identify which standard form the expression resembles — a $\left(1+\Box\right)^{1/\Box}$ shape signals $e$; a $\dfrac{\sin(\Box)}{\Box}$ shape signals Result 9.1(a); a $\dfrac{a^{\Box}-1}{\Box}$ shape signals Result 9.3.
2. Substitute $y$ for the "$\Box$" so the expression matches the standard form exactly, tracking what $y \to$ as $x \to x_0$.
3. Apply the standard limit to the $y$-expression, then (if the exponent or coefficient outside doesn't vanish) combine using the power/product rules from Concept 2. …