A.P.I.D.S. 92

Pada 26hb julai yang lalu, Dr Fadzilah yang merupakan pensyarah bg subjek Algebra & Geometri (FJAX0111) di UM Kuala Lumpur telah hadir ke APIUM Nilam Puri untuk membantu siswa-siswi APIDS yang juga mengambil subjek ini. Kehadiran beliau telah membantu siswa-siswi dalam memahami subjek ini dengan lebih efisyen. Jutaan terima kasih kepada beliau yang sanggup datang dari jauh untuk membantu kami siswa-siswi APIDS disini berkongsi ilmu mungkin dengan cara yang lebih mudah.

Pada 27hb pula, Dr Hj Ismail yang merupakan pensyarah bg subjek Calculus (FJAX0112) di UM Kuala Lumpur juga telah hadir ke sini, APIUM Nilam Puri untuk menambah serba sikit apa yang dirasai perlu ditambah oleh beliau. Jutaan terima kasih juga diucapkan kepada beliau atas kesudiaan untuk berkongsi ilmu dengan kami.

Diharap kepada warga APIDS tidak mensia-siakan ilmu yang dicurahkan oleh mereka malah akan saling membantu antara kita sama sendiri.

p/s=lupa nk tgkap gmbar Dr Hj Ismail..hehe

One of the mysteries of Mathematics seems to be the concept of "infinity", usually denoted by the symbol $\infty$ . So what is $\infty$ ? It is simply a symbol that represents large numbers. Indeed, numbers are of three kinds: large, normal size, and small. The normal size numbers are the ones that we have a clear feeling for. For example, what does a trillion mean? That is a very large number. Also numbers involved in macro-physics are very large numbers. Small numbers are usually used in micro-physics. Numbers like 10^-75 are very small. Being positive or negative has special meaning depending on the problem at hand. The common mistake is to say that $-\infty$ is smaller than 0. While this may be true according to the natural order on the real line in term of sizes, $-\infty$ is big, very big!

So when do we have to deal with $\infty$ and $-\infty$ ? Easy: whenever you take the inverse of small numbers, you generate large numbers and vice-versa. Mathematically we can write this as:

$\begin{displaymath}\frac{1}{0} = \pm \infty\;\;\mbox{and}\;\; \frac{1}{\infty} = 0\;.\end{displaymath}$

Note that the inverse of a small number is a large number. So size-wise there is no problem. But we have to be careful about the positive or negative sign. We have to make sure we know whether a small number is positive or negative. 0+ represents small positive numbers while 0- represents small negative numbers. (Similarly, we will use e.g. 3+ to denote numbers slightly bigger than 3, and 3- to denote numbers slightly smaller than 3.) In other words, being more precise we have

$\begin{displaymath}\frac{1}{0+} = +\infty\;\;\mbox{and}\;\; \frac{1}{0-} = -\infty\;.\end{displaymath}$

Remark. Do not treat $\pm\infty$ as ordinary numbers. These symbols do not obey the usual rules of arithmetic, for instance, $\infty +1=\infty$ , $\infty -1=\infty$ , $2 \cdot\infty=\infty$ , etc.

Example. Consider the function

$\begin{displaymath}f(x) = \frac{1}{x-3}\;\cdot\end{displaymath}$

When $x \rightarrow 3$ , then $x-3 \rightarrow 0$ . So

$\begin{displaymath}\lim_{x \rightarrow 3-} f(x) = \frac{1}{0-} = -\infty\;\;\mbo... ...d}\;\; \lim_{x \rightarrow 3+} f(x) = \frac{1}{0+} = +\infty\;.\end{displaymath}$

Note that when x gets closer to 3, then the points on the graph get closer to the (dashed) vertical line x=3. Such a line is called a vertical asymptote. For a given function f(x), there are four cases, in which vertical asymptotes can present themselves:

(i): $\displaystyle \lim_{x \rightarrow a-} f(x) = -\infty$ ; $\displaystyle \lim_{x \rightarrow a+} f(x) = -\infty$ ;
(ii): $\displaystyle \lim_{x \rightarrow a-} f(x) = -\infty$ ; $\displaystyle \lim_{x \rightarrow a+} f(x) = +\infty$ ;
(iii): $\displaystyle \lim_{x \rightarrow a-} f(x) = +\infty$ ; $\displaystyle \lim_{x \rightarrow a+} f(x) = -\infty$ ;
(iv): $\displaystyle \lim_{x \rightarrow a-} f(x) = +\infty$ ; $\displaystyle \lim_{x \rightarrow a+} f(x) = +\infty$ ;

Next we investigate the behavior of functions when $x \rightarrow \pm\infty$ . We have seen that $\displaystyle \frac{1}{\pm \infty} = 0$ . So for example, we have

$\begin{displaymath}\lim_{x \rightarrow -\infty} \frac{1}{x} = 0\;\;\mbox{and}\;\; \lim_{x \rightarrow +\infty} \frac{1}{x} = 0\;.\end{displaymath}$

In the next example, we show how this result is very useful.

Example. Consider the function

$\begin{displaymath}f(x) = \frac{2x+1}{x-1}\;\cdot\end{displaymath}$

We have

$\begin{displaymath}\frac{2x+1}{x-1} = \frac{2+\displaystyle \frac{1}{x}}{1-\displaystyle \frac{1}{x}}\end{displaymath}$

which implies

$\begin{displaymath}\lim_{x \rightarrow \pm \infty} f(x)= \frac{2+0}{1-0} = 2\;.\end{displaymath}$

Note that when x gets closer to $\pm\infty$ (x gets large), then the points on the graph get closer to the horizontal line y=2. Such a line is called a horizontal asymptote.

In particular, we have

$\begin{displaymath}\lim_{x \rightarrow -\infty} \frac{a}{x^r} = 0\;\;\mbox{and}\;\; \lim_{x \rightarrow +\infty} \frac{a}{x^r} = 0\end{displaymath}$

for any number a, and any positive number r, provided x^r is defined. We also have

$\begin{displaymath}\lim_{x \rightarrow \infty} x^r = \infty\;.\end{displaymath}$

For $-\infty$ , we have to be careful about the definition of the power of negative numbers. In particular, we have

$\begin{displaymath}\lim_{x \rightarrow -\infty} x^n = (-1)^n \infty \end{displaymath}$

for any natural number n.

Example. Consider the function

$\begin{displaymath}f(x) = \frac{2x^4 -3x^2 + 5}{3x^4 + 2x +5}\;\cdot\end{displaymath}$

We have

$\begin{displaymath}\frac{2x^4 -3x^2 + 5}{3x^4 + 2x +5} = \frac{\displaystyle 2 -... ...5}{x^4}}{\displaystyle 3 + \frac{2}{x^3} +\frac{5}{x^4}}\;\cdot\end{displaymath}$

So we have

$\begin{displaymath}\lim_{x \rightarrow \pm \infty} f(x) = \frac{2 -0 + 0}{3 + 0 +0} = \frac{2}{3}\;\cdot\end{displaymath}$

Example. Consider the function

$\begin{displaymath}f(x) = \frac{\sqrt{4x^2 +2}}{3x+1}\;\cdot\end{displaymath}$

We have

$\begin{displaymath}\sqrt{4x^2 + 2} = \sqrt{x^2\left(4 + \frac{2}{x^2}\right)} = \vert x\vert \sqrt{4 + \frac{2}{x^2}}\end{displaymath}$

and then

$\begin{displaymath}\frac{\sqrt{4x^2 +2}}{3x+1} = \frac{\vert x\vert}{x} \frac{\d... ...ystyle \sqrt{4 + \frac{2}{x^2}}}{\displaystyle 3 + \frac{1}{x}}\end{displaymath}$