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One classical application of Taylor expansions is to obtain polynomial equivalents of complicated functions and use them to compute limits.

For example, with Landau notations, we have

$$\begin{array}{rcl} \lim_{x\to 0}\frac{e^x-x-\cos(x)}{x^2} & = & \lim_{x\to 0}\frac{1+x+\frac{x^2}{2}-x-1+\frac{x^2}{2}+o(x^3)}{x^2} \\ & = & \lim_{x\to 0}\frac{x^2+o(x^2)}{x^2}\\ & = & \lim_{x\to 0} 1+o(1)=1 \end{array} $$

But this example can be dealt with using L'Hospital's Rule twice. It seems to me that it would be always the case: since we basically consider ratio of "infinite degree polynomials", we can use repeatly l'Hospital's Rule in order to kill the indetermination.

My question: is there an example where Taylor expansions can be used but not L'Hospital's rule? I guess no so an example where computations with l'Hospital's Rule are awfully complicated but reasonable with Taylor would make me happy.

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Yes, in principle you can always use l'Hopital's rule instead, but in practice there are a few reasons to prefer Taylor series expansions:

  • When you use l'Hopital's rule, you're not only computing Taylor coefficients at the point you care about, but you're also simultaneously computing Taylor coefficients in an interval around the point you care about. This is extra information you don't need and so it's not surprising that the computations are sometimes harder.
  • When you use l'Hopital's rule, you need to check at every step whether you can stop or whether you need to apply l'Hopital's rule again. When you compute Taylor coefficients the coefficients just tell you what's going on, and in particular tell you how many times you would've needed to apply l'Hopital's rule.

This point can be made more explicitly using functions with known Taylor series that are annoying to differentiate and then subtracting off several initial terms. For example, I personally wouldn't want to evaluate

$$\lim_{x \to 0} \frac{\tan x - x + \frac{x^3}{3}}{x^5}$$

by using l'Hopital's rule five times.

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  • $\begingroup$ It should be noted that finding a Taylor series would technically require the computation of five derivatives, making it essentially equivalent to l'Hopital. $\endgroup$
    – user142299
    May 20, 2014 at 16:45
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    $\begingroup$ @Not: sure, if you go from the definition, but of course there are other ways of computing Taylor series. In this case I would instead compute the ratio $\frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} + o(x^7)}{1 - \frac{x^2}{2!} + \frac{x^4}{4!} + o(x^6)}$, and there's a known general expression for the Taylor series of $\tan x$ that was not obtained by differentiating it infinitely many times! $\endgroup$ May 20, 2014 at 16:46
  • $\begingroup$ To reinforce what Qiaochu said, one has all sorts of ways of computing Taylor polynomials (series) from our known toolbox of functions, by multiplying, dividing, composing, using geometric series, etc. Here are two more problems that you'll not enjoy doing with L'Hôpital's rule: $$\lim_{x\to 0} \frac1{\sin^2 x}-\frac 1x,$$ and, of a slighly different flavor, define $$f(x) = \begin{cases} (1+x)^{1/x}, & x\ne 0 \\ e, & x=0\end{cases}.$$ Find $f''(0)$. $\endgroup$ May 20, 2014 at 16:58

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