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Concept 06 / 12Limits, Continuity and Differentiability
Algebra of Limits, Indeterminate Forms and the Limit sin x / x
The sum, product and quotient rules for limits, the seven indeterminate forms, factorisation and rationalisation, the geometric proof that sin x / x tends to 1, and the family of standard trigonometric limits derived from it.
Figure 1
If limx→af(x)=L and limx→ag(x)=M both exist (finite), then the limits of f±g, fg, kf are L±M, LM, kL, and the limit of gf is ML provided M=0. For a continuous function ϕ, limϕ(f(x))=ϕ(L). These rules are what allow 'direct substitution' for polynomials, rational functions (where the denominator is non-zero), and all the standard continuous functions.
The rules have hypotheses. They say nothing when one of the limits does not exist or when substitution produces an indeterminate form. The seven indeterminate forms are
00, ∞∞, 0×∞, ∞−∞, 00, ∞0 and 1∞.
'Indeterminate' means the form alone does not decide the limit: xx, xx2 and x2x are all of type 00 at 0, with limits 1, 0 and 'does not exist'. Such limits must be transformed before evaluating. By contrast, forms such as 01, ∞+∞ or 0∞ are not indeterminate.
Algebraic techniques for 00.
Factorise and cancel the common factor (x−a): limx→2x2−4x3−8=limx+2x2+2x+4=3. Cancelling is legitimate because x=a in the limit.
Rationalise expressions with roots: limx→0x1+x−1=lim1+x+11=21.
Standard algebraic limit: limx→ax−axn−an=nan−1, valid for any rational n (with a>0 where needed). It is simply the derivative of xn at a.
The fundamental trigonometric limit: limx→0xsinx=1, with x in radians. The figure gives the proof for 0<x<2π. In the unit circle, triangle OAP sits inside sector OAP, which sits inside triangle OAT. Their areas are 21sinx, 21x and 21tanx. So sinx<x<tanx. Dividing by sinx>0 and inverting gives cosx<xsinx<1. As x→0+, cosx→1, so the squeeze forces xsinx→1. Since xsinx is even, the left-hand limit agrees. The radian measure is essential: in degrees, limx→0xsinx∘=180π.
Consequences, each derived by rewriting.
limx→0xtanx=1 (write xsinx⋅cosx1).
limx→0x21−cosx=21 (use 1−cosx=2sin22x).
limx→0xsin−1x=limx→0xtan−1x=1 (substitute x=sint or tant).
limx→0bxsinax=ba.
The general principle: if u(x)→0, then u(x)sinu(x)→1. Always match the argument of the sine with the denominator, multiplying and dividing by the right quantity.
For limits at x→a=0, substitute x=a+h so that h→0 and the standard limits apply. For example, limx→ππ−xsinx: with x=π−h, hsin(π−h)=hsinh→1.
Key Takeaways:
- Limit laws need the individual limits to exist; quotient law needs a non-zero denominator limit.
- Indeterminate: 00,∞∞,0⋅∞,∞−∞,00,∞0,1∞.
- limx→ax−axn−an=nan−1.
- xsinx,xtanx,xsin−1x,xtan−1x→1; x21−cosx→21 (radians).
- Match the argument: usinu→1 whenever u→0.
- Shift to h→0 with x=a+h for limits at other points.
- Pitfall: limx→∞xsinx=0, not 1. The standard limit is at x→0.
Brain-Teaser 1: Evaluate limx→0xtan3x1−cos2x.
Solution: 1−cos2x=2sin2x. So the expression is xtan3x2sin2x=2⋅x2sin2x⋅tan3xx. Now tan3xx=31⋅tan3x3x→31. So the limit is 2×1×31=32. Each trigonometric factor is paired with a matching power of x so that it tends to a standard limit.
Brain-Teaser 2: Find limx→1x−1x4−1 and limx→1xn−1xm−1.
Solution: The first is the standard limit with n=4, a=1: 4⋅13=4. For the second, divide numerator and denominator by x−1: (xn−1)/(x−1)(xm−1)/(x−1)→nm. Recognising each piece as a standard limit is faster than polynomial division, especially for fractional exponents.
Brain-Teaser 3: A student claims limx→0xsin(x∘)=1 because 'xsinx→1'. What is the correct value?
Solution: x∘=180πx radians. So xsin(πx/180)=180π⋅πx/180sin(πx/180)→180π. The standard limit requires the argument to be measured in radians, because the area of a sector is 21r2θ only when θ is in radians. This is why calculus is always done in radians: in degrees every derivative of sin and cos would carry an extra factor 180π.
