2. Limits at Infinity and Real Numbers

$Base64 Image$ Hereby L is the candidate value of limits, $\epsilon > 0$ is our degree of precision. Introduce N as a "knob" for tuning $x \geq N$ to achieve the level of precision $\epsilon > 0$ N hereby is defined as when $x \geq N$, $|f(x) - L| \leq \epsilon$. (falls into $[L + \epsilon,L - \epsilon]$) Trivially, as $\epsilon$ decreases, N has to get bigger. Definition of Finite Limit at infinity: Suppose $f: (0,\infty) \rightarrow \mathbb{R}$ and $L \in \mathbb{R}$, then $$ \lim\limits_{x \rightarrow \infty} = L $$ if $\exists N > 0$ for which when $x > N$. $\forall \epsilon > 0$ there exists $|f(x)-L| \leq \epsilon$ Definition of Limit towards infinity: if $\forall k >0$, there exists $N < \infty$ such that when $x > N$, $f(x) > K$, then $$ \lim\limits_{x \rightarrow \infty} = \infty $$ Case where limit does not exist: for $f(x) = x\sin x$, As $x \rightarrow \infty$, proper value of $\epsilon$ and $L$ cannot be determined for limit as $f(x)$ is constantly oscillating. For $\sin x$ it is the same reason that it has no limit.

Limits to Infinity the easy way

Assume $b_m \neq 0$, $a_n \neq 0$ and consider, $$ \begin{matrix} y = \frac{b_mx_m + b_{m-1}x_{m-1}+...+b_0}{a_n x^n + a_{n-1}x^{n-1}+...+a_0} = \frac{x^m}{x^n} \times \frac{b_m+b_{m-1}x^{-1}+...+b_0x^{-m}}{a_n+a_{n-1}x^{-1}+...+a_0x^{-n}}\\ \mathrm{As}\ x \rightarrow \infty,\ \frac{1}{x^n} \rightarrow 0, \mathrm{Therefore}\\ \lim\limits_{x \rightarrow \infty} \frac{b_m+b_{m-1}x^{-1}+...+b_0x^{-m}}{a_n+a_{n-1}x^{-1}+...+a_0x^{-n}} = \frac{b_m}{a_n}\\ \mathrm{Thus}\\ \lim\limits_{x \rightarrow \infty} y = \begin{cases}0&m < n\\\frac{b_m}{a_n}&m=n\\\mathrm{sgn}(\frac{b_m}{a_n})\cdot \infty&m > n \end{cases} \end{matrix} $$ A more complicated example will be: $\lim\limits_{N \rightarrow \infty} \frac{(x+\frac{1}{N})^2-x^2}{\frac{1}{N}} = 2x$

Limits are everywhere!

Real Numbers: $x \in [0,1], x = 0.d_1d_2d_3d_4... = \lim\limits_{N \rightarrow \infty} \sum\limits^N_{i=1} d_i10^{-i} $ Approximate powers of irrational number:: Suppose y is an irrational number, $\{r_i\} \in \mathbb{Q}$ and $|y-r_i| \leq 10^{-i}\ (i\geq 1)$, we have $$ x^y = \lim\limits_{i \rightarrow \infty} x^{r_i} $$ Natural Exponential & Logarithm: $\lim\limits_{n \rightarrow \infty} (1+\frac{1}{n})^n = e $ Growth Rate of functions: $$ \ln x < x^a < a^x < x! < x^x $$ therefore $\lim\limits_{x -> \infty} \frac{a^x}{x^m} = \infty $ and etc.

Algebra on $\infty$

$x + (\pm \infty)$ = $\pm \infty$ $\infty + \infty = \infty$ $\infty - \infty$ is undefined! : indeterminate form