Unit 1, lesson 12

The Formal Definition of Infinite Limits

The precise definitions behind limits equal to positive or negative infinity, at a point, from one side and as x grows without bound, and how to prove or refute them.

  • Theory part 23
  • Practice part 24
  • Limits at Infinity and Infinite Limits

Theory, part 23

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Practice, part 24

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“Grows without bound” sounds clear until you try to prove it. The precise version works like a contest. Someone names a height, as large as they please, and you must show that the function clears that height at every input close enough to the point, the point itself left out. If you can always win, the limit is infinite.

Key terms

M-delta definition at a finite point

Definition Let ff be defined on an open interval containing aa, except possibly at aa. Then lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty means that for every M>0M > 0 there exists δ>0\delta > 0 such that

0<∣x−a∣<δ  ⟹  f(x)>M.0 < \lvert x - a \rvert < \delta \implies f(x) > M.

Allowing every real MM instead of only M>0M > 0 gives the same notion, since clearing a large height also clears every smaller one.

In plain words However high the bar MM is set, every input close enough to aa (and not equal to aa) gives an output above it.

Example We prove that 1/x2→∞1/x^2 \to \infty as x→0x \to 0. Let M>0M > 0 and take δ=1/M\delta = 1/\sqrt{M}. If 0<∣x∣<δ0 < \lvert x \rvert < \delta, then 0<x2<1/M0 < x^2 < 1/M, so 1/x2>M1/x^2 > M. For M=106M = 10^6 this gives δ=0.001\delta = 0.001.

Positive and negative unboundedness

Definition Let ff be defined on an open interval containing aa, except possibly at aa. Then lim⁡x→af(x)=−∞\lim_{x \to a} f(x) = -\infty means that for every N<0N < 0 there exists δ>0\delta > 0 such that

0<∣x−a∣<δ  ⟹  f(x)<N.0 < \lvert x - a \rvert < \delta \implies f(x) < N.

Equivalently, lim⁡x→af(x)=−∞\lim_{x \to a} f(x) = -\infty exactly when lim⁡x→a(−f(x))=∞\lim_{x \to a} \bigl(-f(x)\bigr) = \infty.

In plain words For negative infinity the bar becomes a floor. However deep the floor NN is, the outputs near aa drop below it.

Example We prove lim⁡x→0ln⁡∣x∣=−∞\lim_{x \to 0} \ln \lvert x \rvert = -\infty. Let N<0N < 0 and take δ=eN\delta = e^N, which is positive. If 0<∣x∣<eN0 < \lvert x \rvert < e^N, then ln⁡∣x∣<N\ln \lvert x \rvert < N, because ln⁡\ln is increasing. For N=−10N = -10 this gives δ=e−10≈0.0000454\delta = e^{-10} \approx 0.0000454.

One-sided variants

Definition For ff defined on some interval (a,c)(a, c) with c>ac > a, the statement lim⁡x→a+f(x)=∞\lim_{x \to a^+} f(x) = \infty means that for every M>0M > 0 there exists δ>0\delta > 0 such that

0<x−a<δ  ⟹  f(x)>M.0 < x - a < \delta \implies f(x) > M.

For ff defined on some interval (c,a)(c, a) with c<ac < a, the left-hand version replaces the condition with 0<a−x<δ0 < a - x < \delta. The versions with −∞-\infty replace f(x)>Mf(x) > M with f(x)<Nf(x) < N for every N<0N < 0.

In plain words Keep the height condition and change only which inputs are tested, just as for finite one-sided limits.

Example We prove that 1/x→∞1/x \to \infty as x→0+x \to 0^+. Let M>0M > 0 and take δ=1/M\delta = 1/M. If 0<x<1/M0 < x < 1/M, then 1/x>M1/x > M. On the left, 1/x→−∞1/x \to -\infty as x→0−x \to 0^-. Given N<0N < 0, take δ=−1/N>0\delta = -1/N > 0. If 1/N<x<01/N < x < 0, then 1/x<N1/x < N.

M-N definition at infinity

Definition For ff defined on some interval (c,∞)(c, \infty), the statement lim⁡x→∞f(x)=∞\lim_{x \to \infty} f(x) = \infty means that for every M>0M > 0 there exists a number NN such that

x>N  ⟹  f(x)>M.x > N \implies f(x) > M.

The variants for x→−∞x \to -\infty need ff defined on some interval (−∞,c)(-\infty, c) and use x<Nx < N, and the variants with limit −∞-\infty use f(x)<−Mf(x) < -M in place of f(x)>Mf(x) > M.

In plain words Name a height. Past some point on the xx axis, the graph stays above that height for good.

Example We prove lim⁡x→∞x2=∞\lim_{x \to \infty} x^2 = \infty. Let M>0M > 0 and take N=MN = \sqrt{M}. If x>Mx > \sqrt{M}, then x>0x > 0 and x2>Mx^2 > M. For M=104M = 10^4 this gives N=100N = 100.

Quantifier order

Definition Each definition begins “for every MM there exists δ\delta” (or NN). The input tolerance is chosen after the height, so it may depend on it. Swapping the order to “there exists δ\delta such that for every MM” would require f(x)>Mf(x) > M for every MM at some fixed xx, which no real value can satisfy. The negation of lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty is that there exists M>0M > 0 such that for every δ>0\delta > 0 some xx satisfies

0<∣x−a∣<δandf(x)≤M.0 < \lvert x - a \rvert < \delta \quad \text{and} \quad f(x) \le M.

In plain words The height comes first, then your answer to it. To refute an infinite limit, find one height that the function keeps failing to clear, however close you look.

Example We show that the two-sided limit of 1/x1/x at 00 is not ∞\infty. Take M=1M = 1. For any δ>0\delta > 0 the input x=−δ/2x = -\delta/2 satisfies 0<∣x∣<δ0 < \lvert x \rvert < \delta, and 1/x=−2/δ<0≤11/x = -2/\delta < 0 \le 1. So the bar M=1M = 1 is never cleared on the left side.

Check yourself

  1. For the limit 1/x2→∞1/x^2 \to \infty as x→0x \to 0 and the height M=104M = 10^4, find a δ\delta that works.
Show the answer

The rule from the lesson gives δ=1/104=0.01\delta = 1/\sqrt{10^4} = 0.01. If 0<∣x∣<0.010 < \lvert x \rvert < 0.01, then 0<x2<10−40 < x^2 < 10^{-4} and 1/x2>1041/x^2 > 10^4.

  1. Prove that
lim⁡x→3+1x−3=∞.\lim_{x \to 3^+} \frac{1}{x - 3} = \infty.
Show the answer

Let M>0M > 0 and take δ=1/M\delta = 1/M. If 0<x−3<1/M0 < x - 3 < 1/M, then 1/(x−3)>M1/(x - 3) > M.

  1. Prove that lim⁡x→∞(2x−5)=∞\lim_{x \to \infty} (2x - 5) = \infty.
Show the answer

Let M>0M > 0 and take N=(M+5)/2N = (M + 5)/2. If x>Nx > N, then 2x−5>M+5−5=M2x - 5 > M + 5 - 5 = M.

How this lesson connects