Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 03 / 12Limits, Continuity and Differentiability

L'Hopital's Rule, Series Expansions and the Sandwich Theorem

When and how to apply L'Hopital's rule to 0/0 and infinity/infinity forms, Maclaurin expansions of standard functions for evaluating limits and finding unknown constants, and the sandwich theorem for oscillating and greatest-integer expressions.

Figure 1
L'Hopital's rule. If f(x)→0f(x) \to 0 and g(x)→0g(x) \to 0 as x→ax \to a (or both tend to ±∞\pm\infty), and f,gf, g are differentiable near aa with g′(x)≠0g'(x) \neq 0 near aa, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}, provided the limit on the right exists (or is ±∞\pm\infty). Differentiate the numerator and denominator separately; this is not the quotient rule. The rule can be applied repeatedly as long as the form remains 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. Check the form before each application: applying it to a determinate form gives wrong answers. For example, lim⁡x→0x+1x+2=12\lim_{x \to 0}\frac{x + 1}{x + 2} = \frac{1}{2}, but 'differentiating' gives 11. Other indeterminate forms must be converted first. 0⋅∞0 \cdot \infty: write fg=f1/gfg = \frac{f}{1/g}. ∞−∞\infty - \infty: combine into a single fraction. Exponential forms: take logarithms. When the right-hand limit fails to exist, L'Hopital says nothing. For x+sin⁡xx\frac{x + \sin x}{x} as x→∞x \to \infty, the derivative ratio 1+cos⁡x1 + \cos x oscillates, but the original limit is clearly 11. Use the rule as a tool, not a definition. Series expansions (Maclaurin). Near x=0x = 0: ex=1+x+x22!+x33!+⋯e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots sin⁡x=x−x33!+x55!−⋯\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots cos⁡x=1−x22!+x44!−⋯\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots tan⁡x=x+x33+2x515+⋯\tan x = x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \cdots ln⁡(1+x)=x−x22+x33−⋯\ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots for ∣x∣<1|x| < 1 (1+x)n=1+nx+n(n−1)2x2+⋯(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2}x^{2} + \cdots sin⁡−1x=x+x36+⋯\sin^{-1}x = x + \frac{x^{3}}{6} + \cdots and tan⁡−1x=x−x33+⋯\tan^{-1}x = x - \frac{x^{3}}{3} + \cdots Substituting expansions is often faster than repeated L'Hopital, especially for expressions like tan⁡x−sin⁡xx3\frac{\tan x - \sin x}{x^{3}}, and it is the method of choice when the question asks for constants that make a limit finite. The rule: expand every function up to the order of the lowest non-cancelling power in the denominator. Finding unknown constants. If a limit of the form f(x)xn\frac{f(x)}{x^{n}} is finite, the coefficients of x0,x1,…,xn−1x^{0}, x^{1}, \ldots, x^{n-1} in the expansion of ff must all be zero; the limit is then the coefficient of xnx^{n}. Each vanishing condition gives an equation for the unknowns. The sandwich (squeeze) theorem. If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near aa and lim⁡g=lim⁡h=L\lim g = \lim h = L, then lim⁡f=L\lim f = L. It handles expressions that oscillate or involve [x][x], where algebra and L'Hopital fail. The figure shows xsin⁡1xx\sin\frac{1}{x} trapped between −∣x∣-|x| and ∣x∣|x|: since both bounds tend to 00, so does the function, even though sin⁡1x\sin\frac{1}{x} itself has no limit. The general principle: a bounded function times a function that tends to 00 tends to 00. For the greatest integer function use x−1<[x]≤xx - 1 < [x] \le x; for example, lim⁡x→∞[x]x=1\lim_{x \to \infty}\frac{[x]}{x} = 1 and lim⁡n→∞[x]+[2x]+⋯+[nx]n2=x2\lim_{n \to \infty}\frac{[x] + [2x] + \cdots + [nx]}{n^{2}} = \frac{x}{2}. Key Takeaways: - L'Hopital: 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} only; differentiate top and bottom separately. - Convert 0⋅∞0 \cdot \infty and ∞−∞\infty - \infty to a quotient first. - Expansions: sin⁡x≈x−x36\sin x \approx x - \frac{x^{3}}{6}, tan⁡x≈x+x33\tan x \approx x + \frac{x^{3}}{3}, cos⁡x≈1−x22\cos x \approx 1 - \frac{x^{2}}{2}, ex≈1+x+x22e^{x} \approx 1 + x + \frac{x^{2}}{2}, ln⁡(1+x)≈x−x22\ln(1 + x) \approx x - \frac{x^{2}}{2}. - For a finite lim⁡f(x)xn\lim\frac{f(x)}{x^{n}}: coefficients of lower powers in ff vanish. - Squeeze: bounded ×\times (something →0\to 0) →0\to 0. - x−1<[x]≤xx - 1 < [x] \le x gives squeeze bounds for greatest-integer limits. - Pitfall: applying L'Hopital when the form is not indeterminate. Brain-Teaser 1: Evaluate lim⁡x→0tan⁡x−sin⁡xx3\lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}. Solution: By series: tan⁡x−sin⁡x=(x+x33)−(x−x36)+O(x5)=x32+O(x5)\tan x - \sin x = (x + \frac{x^{3}}{3}) - (x - \frac{x^{3}}{6}) + O(x^{5}) = \frac{x^{3}}{2} + O(x^{5}). So the limit is 12\frac{1}{2}. Alternatively, tan⁡x−sin⁡x=tan⁡x(1−cos⁡x)\tan x - \sin x = \tan x(1 - \cos x), so the expression is tan⁡xx⋅1−cos⁡xx2→1⋅12\frac{\tan x}{x} \cdot \frac{1 - \cos x}{x^{2}} \to 1 \cdot \frac{1}{2}. A common error is to replace tan⁡x\tan x and sin⁡x\sin x both by xx and conclude the limit is 00; substituting first-order approximations inside a difference loses exactly the terms that matter. Brain-Teaser 2: Find aa and bb such that lim⁡x→0x(1+acos⁡x)−bsin⁡xx3=1\lim_{x \to 0}\frac{x(1 + a\cos x) - b\sin x}{x^{3}} = 1. Solution: Expand: x(1+a−ax22)−b(x−x36)+O(x5)=(1+a−b)x+(−a2+b6)x3+O(x5)x(1 + a - \frac{a x^{2}}{2}) - b(x - \frac{x^{3}}{6}) + O(x^{5}) = (1 + a - b)x + (-\frac{a}{2} + \frac{b}{6})x^{3} + O(x^{5}). For a finite limit, the xx coefficient must vanish: 1+a−b=01 + a - b = 0. The limit is then the x3x^{3} coefficient: −a2+b6=1-\frac{a}{2} + \frac{b}{6} = 1. Solving: b=1+ab = 1 + a, so −a2+1+a6=1-\frac{a}{2} + \frac{1 + a}{6} = 1, giving −3a+1+a=6-3a + 1 + a = 6, a=−52a = -\frac{5}{2} and b=−32b = -\frac{3}{2}. Series expansion turns 'find the constants' into two linear equations. Brain-Teaser 3: Evaluate lim⁡x→0x2[1x2]\lim_{x \to 0}x^{2}\left[\frac{1}{x^{2}}\right] and lim⁡x→0x[1x]\lim_{x \to 0}x\left[\frac{1}{x}\right]. Solution: For any real tt, t−1<[t]≤tt - 1 < [t] \le t. With t=1x2t = \frac{1}{x^{2}} and multiplying by x2>0x^{2} > 0: 1−x2<x2[1x2]≤11 - x^{2} < x^{2}[\frac{1}{x^{2}}] \le 1. Both bounds tend to 11, so the limit is 11. For the second, as x→0+x \to 0^{+} the same argument gives 11. As x→0−x \to 0^{-}, multiplying 1x−1<[1x]≤1x\frac{1}{x} - 1 < [\frac{1}{x}] \le \frac{1}{x} by x<0x < 0 reverses the inequalities: 1≤x[1x]<1−x1 \le x[\frac{1}{x}] < 1 - x, which also squeezes to 11. So both limits are 11. Multiplying an inequality by a negative quantity reverses it, and forgetting this is the usual error on the left side.