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Limits at infinity

12 November, 2015 - 15:03

The definition of the limit in terms of infinitesimals extends immediately to limiting processes where x gets bigger and bigger, rather than closer and closer to some finite value. For example, the function 3+1/x clearly gets closer and closer to 3 as x gets bigger and bigger. If a is an infinite number, then the definition says that evaluating this expression at a+dx, where dx is infinitesimal, gives a result whose standard part is 3. It doesn’t matter that a happens to be infinite, the definition still works. We also note that in this example, it doesn’t matter what infinite number a is; the limit equals 3 for any infinite a. We can write this fact as
\lim_{x\rightarrow \infty}\left ( 3+\frac{1}{x} \right )=3

where the symbol \infty is to be interpreted as “nyeah nyeah, I don’t even care what infinite number you put in here, I claim it will work out to 3 no matter what.” The symbol \infty is not to be interpreted as standing for any specific infinite number. That would be the type of fallacy that lay behind the bogus proof in Safe use of infinitesimals that 1 = 1/2, which assumed that all infinities had to be the same size.

A somewhat different example is the arctangent function. The arctangent of 1000 equals approximately 1.5698, and inputting bigger and bigger numbers gives answers that appear to get closer and closer to\pi/2\approx 1.5707. But the arctangent of -1000 is approximately -1.5698, i.e., very close to-\pi/2. From these numerical observations, we conjecture that\lim_{x \to a} \textrm{tan }^{-1}x

equals \pi/2 for positive infinite a, but -\pi/2 for negative infinite a. It would not be correct to write
\begin{matrix} \lim_{x\rightarrow \infty} \textrm{tan}^{-1}x=\frac{\pi}{2} & \textrm{[wrong]} \end{matrix}

because it does matter what infinite number we pick. Instead we write

\begin{align*} \lim_{x\rightarrow +\infty}\textrm{tan }^{-1} x&=\frac{\pi}{2} \\ \lim_{x\rightarrow -\infty} \textrm{tan }^{-1} x &= -\frac{\pi}{2} \end{align*}

Some expressions don't have this kind of limit at all. For example, if you take the sines of big numbers like a thousand, a million, etc., on your calculator, the results are essentially random numbers lying between 1 and 1. They don't settle down to any particular value, because the sine function oscillates back and forth forever. To prove formally that \textrm{lim}_{x\rightarrow +\infty} \textrm{sin }x is undefined, consider that the sine function, defined on the real numbers, has the property that you can always change its result by at least 0.1 if you add either 1.5 or 1.5 to its input. For example, \textrm{sin}(.8)\approx 0.717, and \textrm{sin}(.8-1.5)\approx 0.644. Applying the transfer principle to this statement, we find that the same is true on the hyperreals. Therefore there cannot be any value l that differs infinitesimally from \textrm{sin }a for all positive infinite values of a.

Often we're interested in finding the limit as x approaches infinity of an expression that is written as an indeterminate form like H/K, where both H and K are infinite.

Example

Evaluate the limit

\lim_{x\rightarrow \infty}\frac{2x+7}{x+8686}

Intuitively, if x gets large enough the constant terms will be negligible, and the top and bottom will be dominated by the 2x and x terms, respectively, giving an answer that approaches 2.

One way to verify this is to divide both the top and the bottom by x, giving

\frac{2+\frac{7}{x}}{1+\frac{8686}{x}}

If x is infinite, then the standard part of the top is 2, the standard part of the bottom is 1, and the standard part of the whole thing is therefore 2.

Another approach is to use l’Hopital’s rule. The derivative of the top is 2, and the derivative of the bottom is 1, so the limit is 2/1=2.