Limits that arise frequently

We continue our presentation of basic stuff from Calculus and Analytic Geometry, G B Thomas and Finney, Ninth Edition. My express purpose in presenting these few proofs is to emphasize that Calculus, is not just a recipe of calculation techniques. Or, even, a bit further, math is not just about calculation. I have a feeling that such thinking nurtured/developed at a young age, (while preparing for IITJEE Math, for example) makes one razor sharp.

We verify a few famous limits.

Formula 1:

If |x|<1, \lim_{n \rightarrow \infty}x^{n}=0

We need to show that to each \in >0 there corresponds an integer N so large that |x^{n}|<\in for all n greater than N. Since \in^{1/n}\rightarrow 1, while |x|<1. there exists an integer N for which \in^{1/n}>|x|. In other words,

|x^{N}|=|x|^{N}<\in. Call this (I).

This is the integer we seek because, if |x|<1, then

|x^{n}|<|x^{N}| for all n>N. Call this (II).

Combining I and II produces |x^{n}|<\in for all n>N, concluding the proof.

Formula II:

For any number x, \lim_{n \rightarrow \infty}(1+\frac{x}{n})^{n}=e^{x}.

Let a_{n}=(1+\frac{x}{n})^{n}. Then, \ln {a_{n}}=\ln{(1+\frac{x}{n})^{n}}=n\ln{(1+\frac{x}{n})}\rightarrow x,

as we can see by the following application of l’Hopital’s rule, in which we differentiate with respect to n:

\lim_{n \rightarrow \infty}n\ln{(1+\frac{x}{n})}=\lim_{n \rightarrow \infty}\frac{\ln{(1+x/n)}}{1/n}, which in turn equals

\lim_{n \rightarrow \infty}\frac{(\frac{1}{1+x/n}).(-\frac{x}{n^{2}})}{-1/n^{2}}=\lim_{n \rightarrow \infty}\frac{x}{1+x/n}=x.

Now, let us apply the following theorem with f(x)=e^{x} to the above:

(a theorem for calculating limits of sequences) the continuous function theorem for sequences:

Let a_{n} be a sequence of real numbers. If \{a_{n}\} be a sequence of real numbers. If a_{n} \rightarrow L and if f is a function that is continu0us at L and defined at all a_{n}, then f(a_{n}) \rightarrow f(L).

So, in this particular proof, we get the following:

(1+\frac{x}{n})^{n}=a_{n}=e^{\ln{a_{n}}}\rightarrow e^{x}.

Formula 3:

For any number x, \lim_{n \rightarrow \infty}\frac{x^{n}}{n!}=0

Since -\frac{|x|^{n}}{n!} \leq \frac{x^{n}}{n!} \leq \frac{|x|^{n}}{n!},

all we need to show is that \frac{|x|^{n}}{n!} \rightarrow 0. We can then apply the Sandwich Theorem for Sequences (Let \{a_{n}\}, \{b_{n}\} and \{c_{n}\} be sequences of real numbers. if a_{n}\leq b_{n}\leq c_{n} holds for all n beyond some index N, and if \lim_{n\rightarrow \infty}a_{n}=\lim_{n\rightarrow \infty}c_{n}=L,, then \lim_{n\rightarrow \infty}b_{n}=L also) to  conclude that \frac{x^{n}}{n!} \rightarrow 0.

The first step in showing that |x|^{n}/n! \rightarrow 0 is to choose an integer M>|x|, so that (|x|/M)<1. Now, let us the rule (formula 1, mentioned above), so we conclude that:(|x|/M)^{n}\rightarrow 0. We then restrict our attention to values of n>M. For these values of n, we can write:

\frac{|x|^{n}}{n!}=\frac{|x|^{n}}{1.2 \ldots M.(M+1)(M+2)\ldots n}, where there are (n-M) factors in the expression (M+1)(M+2)\ldots n, and

the RHS in the above expression is \leq \frac{|x|^{n}}{M!M^{n-M}}=\frac{|x|^{n}M^{M}}{M!M^{n}}=\frac{M^{M}}{M!}(\frac{|x|}{M})^{n}. Thus,

0\leq \frac{|x|^{n}}{n!}\leq \frac{M^{M}}{M!}(\frac{|x|}{M})^{n}. Now, the constant \frac{M^{M}}{M!} does not change as n increases. Thus, the Sandwich theorem tells us that \frac{|x|^{n}}{n!} \rightarrow 0 because (\frac{|x|}{M})^{n}\rightarrow 0.

That’s all, folks !!


Nalin Pithwa.

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