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Business Mathematics and Basic Statistics · Ch 10 — Limits and Derivatives

Exponential and Logarithmic Standard Limits

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Exponential and Logarithmic Standard Limits

Two more standard limits — both again of the indeterminate 00\tfrac00 form on direct substitution — are used constantly once exponential and logarithmic functions enter business-mathematics problems (compound growth, continuous interest, and, later in this chapter, differentiation of exe^x and ln⁡x\ln x):

lim⁡x→0ex−1x=1lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{e^{x}-1}{x} = 1 \qquad\qquad \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1

Both say the same kind of thing as §4's formulae: the expression is undefined at x=0x=0 itself (00\tfrac00), but the function it defines approaches exactly 11 as xx gets arbitrarily close to 00.

How to apply them to a scaled argument (the most common exam variant): for lim⁡x→0ekx−1x\displaystyle\lim_{x\to0}\frac{e^{kx}-1}{x}, substitute u=kxu=kx (so u→0u\to0 as x→0x\to0, and x=u/kx=u/k):

lim⁡x→0ekx−1x=lim⁡u→0eu−1u/k=k⋅lim⁡u→0eu−1u=k⋅1=k\lim_{x\to0}\frac{e^{kx}-1}{x} = \lim_{u\to0}\frac{e^{u}-1}{u/k} = k\cdot\lim_{u\to0}\frac{e^u-1}{u} = k\cdot 1 = k

The same substitution technique gives lim⁡x→0ln⁡(1+kx)x=k\displaystyle\lim_{x\to0}\frac{\ln(1+kx)}{x} = k.

Note

Where These Formulae Come From, Informally …

Definition 6Exponential standard limit

lim⁡x→0ex−1x=1\lim_{x\to0}\dfrac{e^x-1}{x}=1. For a scaled argument, $\lim_{x\to0}\dfrac{e^ …

Definition 7Logarithmic standard limit

lim⁡x→0ln⁡(1+x)x=1\lim_{x\to0}\dfrac{\ln(1+x)}{x}=1. For a scaled argument, $\lim_{x\to0}\dfrac{\l …