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)≤f(x)≤h(x) for all x near x0 (except possibly at x0 itself), and if
limx→x0g(x)=limx→x0h(x)=l,
then limx→x0f(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 x→0limx2sinx21=0, note that sin(⋅) is always between −1 and 1, so −x2≤x2sinx21≤x2. Since both −x2 and x2 tend to 0 as x→0, the Sandwich Theorem forces the middle expression to 0 as well — even though limx→0sinx21 on its own does not exist (it oscillates wildly), so the product rule alone could never have been applied directly.
This is exactly why the Sandwich Theorem is indispensable rather than a curiosity: whenever one factor oscillates without a limit but is bounded, and the other factor is squeezed to zero, the ordinary product law (Concept 2) is not applicable — you need the sandwich.
The two flagship trigonometric limits
Result 9.1.
(a)limθ→0θsinθ=1(b)limθ→0θ1−cosθ=0
Part (a) is proved geometrically by sandwiching θsinθ between cosθ 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 θ→0, so must θsinθ. Part (b) follows algebraically from (a) by writing 1−cosθ=2sin22θ 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: x→0limsinx=0, obtained from the sandwich −∣x∣≤sinx≤∣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):
limx→0xex−1=1limx→0xax−1=loga (a>0)limx→0xlog(1+x)=1
limx→0xsin−1x=1limx→0xtan−1x=1
And the three equivalent forms of the number e as a limit:
limx→∞(1+x1)x=elimx→0(1+x)1/x=elimx→∞(1+xk)x=ek
e is a transcendental number — it never satisfies any polynomial equation with rational coefficients. That's part of why it shows up as a genuinely new limiting constant here rather than something expressible in simpler closed form.
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=(some expression in x) chosen so that y→0 (or y→∞) exactly when x does. The book's worked examples all follow this shape:
- Spot the shell. Identify which standard form the expression resembles — a (1+□)1/□ shape signals e; a □sin(□) shape signals Result 9.1(a); a □a□−1 shape signals Result 9.3.
- Substitute y for the "□" so the expression matches the standard form exactly, tracking what y→ as x→x0.
- 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. …