## 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 !!

Aufwiedersehen,

Nalin Pithwa.

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