扩展：由泰勒近似得到的极限值

In this section, we will use Taylor approximations to calculate the limits of rational indeterminate forms. Remember that a Taylor polynomial can approximate a function very accurately within a certain neighborhood of a point, say $x = a$, and a limit is only concerned with the behavior of a function around an arbitrary point $x = a$. Therefore, it seems logical to use Taylor polynomials to our advantage in solving tough limit questions.

Let's start with a basic example $\lim_{x \to 0} \frac{\text{sin} x}{x}$. To do this job, one's first reaction may be to use L'Hopital's rule. This may work; however, we will proceed via Taylor polynomials. As discussed earlier, the following Taylor approximation is valid. $$\text{sin}x = x - \frac{x^3}{6} + \frac{x^5}{120} \cdots \; \;(1)$$

Now, using equation 1, we can substitute into our original problem. $$\lim_{x \to 0} \frac{\text{sin} x}{x}$$ $$\leftrightarrow \lim_{x \to 0} \frac{x - \frac{x^3}{6} + \frac{x^5}{120} \cdots}{x}$$

Note that we can split this limit up into the following: $$\lim_{x \to 0} \frac{x}{x} + \frac{x^3}{6x} + \frac{x^5}{120x} \cdots$$

In this case, the limit of $\frac{x}{x}$ is one, as it is a constant. $\frac{x^3}{6x}$ simplifies to $\frac{x^2}{6}$, whose limit as x goes to 0 is 0. This is true for all succeeding terms, as the degree of the terms becomes greater.

Note that this method is useful especially if you have a quick recall of Taylor series and Taylor polynomials. After one remembers the Taylor approximation for a function, the problem reduces to one back from precalculus.

Finally, we will end with a question.

Submitted by Sujay Kazi

(2015 ASMT, Calculus #4) Given that $\alpha$ and $\beta$ are positive real numbers, compute the following limit (where it exists and is nonzero) in terms of $\alpha$ and $\beta$: $$\lim_{x\rightarrow 0^{+}} \frac{\sin x^{\alpha}}{\cos x^{\beta} -1}.$$

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Hint

Use the Taylor expansions for $\sin x$ and $\cos x$ centered at $x=0$: $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$$ $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$$

[Click Continue for the answer.]

Answer

$-2$

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Solution

Using the Taylor expansions for $\sin x$ and $\cos x$ centered at $x=0$, we see that: $$\sin x^{\alpha} = x^{\alpha} - \frac{x^{3\alpha}}{6} + \frac{x^{5\alpha}}{5!} - \frac{x^{7\alpha}}{7!} + ...$$ $$\cos x^{\beta} - 1 = -\frac{x^{2\beta}}{2} + \frac{x^{4\beta}}{4!} - \frac{x^{6\beta}}{6!} + \frac{x^{8\beta}}{8!} - ...$$ As $x$ approaches $0$ from the positive side, all of the terms in each series except the first become insignificant. Thus, it suffices to compare: $$\sin x^{\alpha} \approx x^{\alpha}$$ $$\cos x^{\beta} - 1 \approx -\frac{x^{2\beta}}{2}$$ For the desired limit to exist and be nonzero, the exponents of $x$ must match, so $\alpha = 2\beta$. A comparison of the two approximations shows that the limit must equal $-2$, which is the answer.