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Concept 02 / 12Limits, Continuity and Differentiability
Exponential and Logarithmic Limits and the 1-to-the-Infinity Form
The number e as the limit of (1 + 1/n)^n, the standard limits (e^x - 1)/x, (a^x - 1)/x and ln(1 + x)/x, the shortcut e^(lim g(f - 1)) for 1^infinity forms, and handling 0^0 and infinity^0 by taking logarithms.
Figure 1
The number e is defined by the limit
e=limn→∞(1+n1)n=limx→0(1+x)1/x.
The figure shows the sequence (1+n1)n rising towards e≈2.71828. This is the prototype of the indeterminate form 1∞: the base tends to 1 and the exponent to infinity, and the two effects balance to give a finite number that is neither 1 nor ∞. The compound-interest interpretation makes it concrete: 100% annual interest compounded n times a year multiplies money by (1+n1)n, which approaches e as compounding becomes continuous.
Standard limits, all with x→0:
limxex−1=1.
limxax−1=lna for a>0 (write ax=exlna).
limxln(1+x)=1 (take the logarithm of (1+x)1/x→e).
limxloga(1+x)=lna1.
limx(1+x)n−1=n for any real n.
As with xsinx, each generalises by replacing x with any u(x)→0: ueu−1→1 and uln(1+u)→1.
The 1∞ shortcut. Suppose f(x)→1 and g(x)→∞. Write fg=[1+(f−1)]g={[1+(f−1)]1/(f−1)}g(f−1). The inner bracket tends to e, so
limf(x)g(x)=elimg(x)[f(x)−1].
This turns every 1∞ problem into a single ordinary limit of g(f−1), usually of type ∞×0 that simplifies easily. For example, limx→0(cosx)1/x2=elimx2cosx−1=e−1/2.
The forms 00 and ∞0. Take logarithms: if y=fg, then lny=glnf, which is of type 0×∞. Evaluate limlny=L (often by rewriting as a quotient and using L'Hopital's rule or a standard limit), then the answer is eL. For example, limx→0+xx: lny=xlnx→0 (since lnx grows more slowly than x1), so the limit is e0=1. And limx→∞x1/x: lny=xlnx→0, so the limit is 1.
Growth rates at infinity. As x→∞, exponentials beat powers and powers beat logarithms: for any n>0 and a>1, axxn→0 and x(lnx)n→0. As x→0+, xnlnx→0 for n>0. These facts decide most ∞∞ and 0⋅∞ limits involving mixed function types.
Key Takeaways:
- e=lim(1+n1)n=limx→0(1+x)1/x.
- xex−1→1; xax−1→lna; xln(1+x)→1 as x→0.
- 1∞: limfg=elimg(f−1).
- 00, ∞0: take logs, find the limit L of glnf, answer eL.
- Growth: lnx≪xn≪ax as x→∞.
- limx→0+xx=1.
- Pitfall: declaring 1∞=1. The form is indeterminate.
Brain-Teaser 1: Evaluate limx→∞(x−1x+3)x.
Solution: The base tends to 1 and the exponent to ∞. Using the shortcut: g(f−1)=x(x−1x+3−1)=x⋅x−14→4. So the limit is e4. The shortcut reduces a hard-looking power to a one-line rational limit.
Brain-Teaser 2: Find limx→0tanx−xetanx−ex.
Solution: Factor out ex: tanx−xex(etanx−x−1). Let u=tanx−x→0. Then ueu−1→1 and ex→1, so the limit is 1. No expansion of tanx is needed, because the difference tanx−x appears in both places. Spotting a hidden standard form with a composite argument is the key skill.
Brain-Teaser 3: Evaluate limx→0(31x+2x+3x)1/x.
Solution: At x=0 the base is 1, so this is 1∞. The exponent of e is limx1(31x+2x+3x−1)=lim31(x1x−1+x2x−1+x3x−1)=31(ln1+ln2+ln3)=3ln6. So the limit is eln6/3=61/3, the geometric mean of 1,2,3. In general, this limit gives the geometric mean of the bases, a pleasing link between limits and means.
