example of computing limits using Taylor expansion

Problem.

Evaluate

$$\underset{x\to -\mathrm{\infty }}{lim}\left(x+\sqrt{x^2+23x}\right).$$

In beginners’ courses in calculus, one is usually told torationalize the above expression by multiplying by $(x-\sqrt{x^2+23x})/(x-\sqrt{x^2+23x})$.This approach is somewhat dissatisfying, because it dependson a specific algebraic trick that works only for square roots— for example, it would have been more difficult or impossible to rationalize, if instead, we had a cube root or even a transcendental function —and this trick does not appeal to our intuition that$\sqrt{x^2+23x}$ should be approximately$|x|$ for large $|x|$.Fortunately, there is another approach thatexploits the analytic properties of the functions involved.

L’Hôpital’s Rule (http://planetmath.org/LHpitalsRule) is one analytic approach,but in many cases, using Taylor expansion is even easierand straightforward. Essentially, Taylor expansion approximatescomplicated functions by polynomials, whose limits are easy to evaluate.We illustrate the method below.

First rewrite

$$\underset{x\to -\mathrm{\infty }}{lim}\left(x+\sqrt{x^2+23x}\right)=\underset{x\to +\mathrm{\infty }}{lim}\left(\sqrt{x^2-23x}-x\right),$$

so that we do not have to worry about the pesky negatives any more.Then, with the help of the binomial formula:

$${(1+y)}^{1/2}=1+\frac{1}{2}y+o(y),\text{as }y\to 0\text{,}$$

(“$o$” is Landau notation)we obtain:

$\sqrt{x^2-23x}-x$$=x\left(\sqrt{1-{\displaystyle \frac{23}{x}}}-1\right)$

$=x\left(1-{\displaystyle \frac{1}{2}}{\displaystyle \frac{23}{x}}+o\left({\displaystyle \frac{23}{x}}\right)-1\right),\text{as }x\to \mathrm{\infty }\text{ (so }y=-\frac{23}{x}\to 0\text{)}$

$=-{\displaystyle \frac{23}{2}}+xo\left({\displaystyle \frac{23}{x}}\right)$

$=-{\displaystyle \frac{23}{2}}+o(1).$

Therefore

$$\underset{x\to -\mathrm{\infty }}{lim}\left(x+\sqrt{x^2+23x}\right)=\underset{x\to \mathrm{\infty }}{lim}\left(\sqrt{x^2-23x}-x\right)=-\frac{23}{2}.$$

Problem.

Evaluate

$$\underset{x\to 0}{lim}\left(\frac{1}{\ln (1+x)}-\frac{1}{\tan x}\right).$$

This example is admittedly artificial; it was made to be annoying to solve usingL’Hôpital’s Rule alone, but much simpler if one knows how to usethe Taylor expansions:

$\ln (1+x)$$=x-{\displaystyle \frac{1}{2}}{x}^2+o(x^2),$as $x\to 0$, and

$\tan x$$=x+o(x^2),$as $x\to 0$.

So we compute:

${\displaystyle \frac{1}{\ln (1+x)}}-{\displaystyle \frac{1}{\tan x}}={\displaystyle \frac{\tan x-\ln (1+x)}{\ln (1+x)\tan x}}$$={\displaystyle \frac{\left(x+o(x^2)\right)-\left(x-\frac{1}{2}{x}^2+o(x^2)\right)}{\ln (1+x)\tan x}}$

$={\displaystyle \frac{\frac{1}{2}{x}^2+o(x^2)}{\left(x+o(x)\right)\left(x+o(x)\right)}}$

$={\displaystyle \frac{\frac{1}{2}+o(1)}{\left(1+o(1)\right)\left(1+o(1)\right)}}$

Therefore,

$$\underset{x\to 0}{lim}\left(\frac{1}{\ln (1+x)}-\frac{1}{\tan x}\right)=\frac{1}{2}.$$

The reader might reasonably ask howdid we know the right number of terms to use in the Taylor expansions.The answer is to guess.This is not as problematic as it sounds. First, there is no harmin using more terms than necessary in the expansion (only that there is more writing).And if we used too few terms, we would know when we later encounter indeterminate forms such as $o(1)/x$ (as $x\to 0$) in our derivations. If that happens,it is not hard to go back and add the needed terms.

Notice that all the essential information to evaluate the limit is contained in the first few derivatives ofthe functions involved at particular points — in the above example,only at $x=0$.This information can be obtained by manipulating series,unlike L’Hôpital’s Rulewhich necessitates computing the derivative functions at all points.So even monstrous expressions like this one is tractablewith Taylor expansion:

$$\underset{x\to 0}{lim}\left(\frac{1}{\ln (1+\tan (\sin x))}-\frac{1}{e^{\sin (\tan x)}-1}\right).$$

On the other hand, there definitely are situations whereL’Hôpital’s Rule works but Taylor expansion does not: for instance,

$$\underset{x\to 0^{+}}{lim}x\ln x=\underset{x\to 0^{+}}{lim}\frac{x}{1/\ln x},$$

because $1/\ln x$ cannot be expanded in a Taylor series about $x=0$.

Problem.

Here is a problem involving a different sort of limit:does the following series converge?

$$\sum _{n=1}^{\mathrm{\infty }}\arctan (n^{-1})$$

Our intuition suggests that it does not, because$\arctan (n^{-1})$ should be approximately $n^{-1}$, and ${\sum }_n{n}^{-1}$ diverges.However, the standard comparison test does not workbecause $\arctan x\le x$ (for $x\ge 0$) has the inequality in the wrong direction.But with Taylor expansion the solution is a snap.By expanding

$$\arctan x=x+O(x^3),\text{as }x\to 0\text{.}$$

and summing both sides, we get

$$\sum _{n=1}^{\mathrm{\infty }}\arctan (n^{-1})=\sum _{n=1}^{\mathrm{\infty }}{n}^{-1}+\sum _{n=1}^{\mathrm{\infty }}O(n^{-3}).$$

As ${\sum }_{n}O(n^{-3})$ converges (being dominated by $C{\sum }_n{n}^{-3}$ for some constant $C$), ${\sum }_n\arctan (n^{-1})$ must diverge (to $\mathrm{\infty }$).

(Of course, this problem could be solved by using the integral test,but who really wants to integrate $\int \arctan (1/x)𝑑x$?)