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Mathematics · Ch 9 — Differential Calculus – Limits and Continuity

Applications of limits

9.2.7

Applications of limits

Limits at infinity and one-sided limits are not merely abstract exercises — they routinely answer a natural real-world question: "what is the extreme (maximum or minimum, long-run or boundary) behaviour of this quantity?"

Maximum processing rate (Example 9.25). If the rate rr at which the liver removes alcohol from the bloodstream, as a function of the blood alcohol concentration xx, follows r(x)=αxβ+xr(x)=\dfrac{\alpha x}{\beta+x} for positive constants α,β\alpha,\beta, then since rr increases with xx, the maximum possible rate is obtained by letting concentration grow without bound:

lim⁡x→∞r(x)=lim⁡x→∞αxβ+x=lim⁡x→∞α1+β/x=α\lim_{x\to\infty} r(x)=\lim_{x\to\infty}\frac{\alpha x}{\beta+x}=\lim_{x\to\infty}\frac{\alpha}{1+\beta/x}=\alpha

(dividing numerator and denominator by xx — the technique of Section 9.2.5). So the liver's processing rate can never exceed α\alpha, however high the concentration climbs — a genuine physical ceiling read directly off a limit at infinity.

A one-sided limit forced by physics (Example 9.26). Einstein's relativistic mass formula m=m01−v2/c2m=\dfrac{m_0}{\sqrt{1-v^2/c^2}} (where cc is the speed of light) is only physically meaningful for v<cv<c, so asking what happens "as vv approaches cc" only makes sense as a left-hand limit, v→c−v\to c^-; a genuine two-sided limit is not even defined, since v>cv>c would make the quantity under the square root negative. As v→c−v\to c^-, the denominator 1−v2/c2→0+\sqrt{1-v^2/c^2}\to0^+ (through positive values only), so m→∞m\to\infty: mass grows without bound as an object's speed nears the speed of light — exactly the physical statement that no massive object can actually be pushed to reach cc.

Terminal velocity (Example 9.27). A falling object's velocity model r(t)=32k⋅1−e−32kt1+e−32ktr(t)=\dfrac{32}{k}\cdot\dfrac{1-e^{-32kt}}{1+e^{-32kt}} involves an exponential term e−32kte^{-32kt} that vanishes as t→∞t\to\infty (for k>0k>0), so the long-run ("terminal") velocity is found simply by letting that exponential term tend to 00 inside the formula:

lim⁡t→∞r(t)=32k⋅1−01+0=32k ft/sec—\lim_{t\to\infty} r(t)=\frac{32}{k}\cdot\frac{1-0}{1+0}=\frac{32}{k}\ \text{ft/sec} —

the speed at which air resistance and gravity come into balance. …