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mathsbrain · Mar 14, 2014

1. Where can you buy 4unit textbook by someone called komoroni

2. How do you sketch y=x/((e^X) -1) without lhopital

Sy123 · Mar 14, 2014

$$Consider the differentiation of$ \ f(x) = e^x \ $by first principles$ \\ \\ (e^x)' = \lim_{h\to 0} \frac{e^{x+h}-e^x}{h} = e^x \lim_{h \to 0} \frac{e^h-1}{h} \\ \\ (e^x)'=e^x \\ \\ $therefore$ \ e^x\lim_{h \to 0} \frac{e^{h}-1}{h} = e^x \ \rightarrow \ \lim_{h \to 0} \frac{e^h-1}{h} = 1 \\ \\ \Rightarrow \lim_{x \to 0} \frac{x}{e^x-1} = 1$

$f'(x) = \frac{(1-x)e^x - 1}{(e^x-1)^2} \\ \\ $find values of$ \ x \ $for$ \ (1-x)e^x > 1$

$$By graping$ \ (1-x)e^x \ $it is found that there are no real$ \ x \ $for$ \ (1-x)e^x > 1$

$\therefore \ f'(x) \leq 0$

$$It is clear that$ \ \lim_{x \to \infty} f(x) = 0$

$$No stationary point, hence graph looks like$ \ y=e^{-x} \ $with hollow circle at$ \ (0,1)$

mathsbrain · Mar 14, 2014

Sy123 said:
$$Consider the differentiation of$ \ f(x) = e^x \ $by first principles$ \\ \\ (e^x)' = \lim_{h\to 0} \frac{e^{x+h}-e^x}{h} = e^x \lim_{h \to 0} \frac{e^h-1}{h} \\ \\ (e^x)'=e^x \\ \\ $therefore$ \ e^x\lim_{h \to 0} \frac{e^{h}-1}{h} = e^x \ \rightarrow \ \lim_{h \to 0} \frac{e^h-1}{h} = 1 \\ \\ \Rightarrow \lim_{x \to 0} \frac{x}{e^x-1} = 1$

$f'(x) = \frac{(1-x)e^x - 1}{(e^x-1)^2} \\ \\ $find values of$ \ x \ $for$ \ (1-x)e^x > 1$

$$By graping$ \ (1-x)e^x \ $it is found that there are no real$ \ x \ $for$ \ (1-x)e^x > 1$

$\therefore \ f'(x) \leq 0$

$$It is clear that$ \ \lim_{x \to \infty} f(x) = 0$

$$No stationary point, hence graph looks like$ \ y=e^{-x} \ $with hollow circle at$ \ (0,1)$

Why is limit as x -> infinity that y goes to 0? How do you explain this without using l'hopital's rule?

QZP · Mar 14, 2014

mathsbrain said:
Why is limit as x -> infinity that y goes to 0? How do you explain this without using l'hopital's rule?

It's quite intuitive. Or use a calculator

braintic · Mar 14, 2014

To the best of my knowledge, Komaromi (the correct spelling) did not publish a textbook on the entire 4 unit course.
He produced separate books on Volumes, Graphs and Circular Motion (but not resisted motion). I believe he also had a book on Conics, though I have never seen it. The first two are spiral-bound books using the old foolscap paper size. I don't think they have been available for a number of years. I have physical copies of the first three, but no digital copies.

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Regarding l'Hopital's rule .... at a recent conference on the marking of the 2013 HSC maths papers, they made it clear that any results that are beyond the NSW HSC syllabus (such as l'Hopital's rule) must be FULLY explained in order to use them. If you use l'Hopital's rule without any explanation you will get zero marks.

Sy123 · Mar 15, 2014

mathsbrain said:
Why is limit as x -> infinity that y goes to 0? How do you explain this without using l'hopital's rule?

As x goes to infinity, e^x dominates the x on top, in heuristic terms e^x goes to infinity "faster" than x

Sy123 · Mar 15, 2014

Sy123 said:
As x goes to infinity, e^x dominates the x on top, in heuristic terms e^x goes to infinity "faster" than x

As for a rigorous proof, we note that

$\\ $For large enough$ \ x \ $note that$ \\ \\ 0 < \frac{x}{e^x-1} < \frac{1}{x} \\ \\ $The upper bound can be proven with calculus, as with the lower bound$ \\ \\ $Hence by taking$ \ x \ $to infinity$ \ 0 < \lim_{x \to \infty} \frac{x}{e^x-1} < \lim_{x \to \infty} \frac{1}{x} = 0 \\ \\ $Hence$ \ \lim_{x \to \infty} \frac{x}{e^x-1} = 0$

Sy123 · Mar 15, 2014

Alternatively for the original limit as x goes to zero, we can see that

$\frac{x}{e^x-1} < 1 \\ $Since$ \ e^x > x+1 \ $proven by calculus or otherwise$ \\ \\ $And for small enough positive$ \ x , \ \ \ \ e^x < x+2 \ \ \ $proven with calculus or otherwise$ \\ \\ $Hence for small enough$ \ x , \ \ \ \frac{x}{e^x-1} > \frac{x}{x+1} \\ \\ $Hence for small enough$ x , \ \ \ \frac{x}{x+1} < \frac{x}{e^x-1} < 1 \\ \\ $Therefore$ \ 1=\lim_{x \to 0} \frac{x}{x+1} < \lim_{x \to 0} \frac{x}{e^x-1} < 1 \\ \\ \therefore \ \lim_{x \to 0} \frac{x}{e^x-1} = 1$

Trebla · Mar 16, 2014

mathsbrain said:
2. How do you sketch y=x/((e^X) -1) without lhopital

If you're just sketching don't you just apply the principles of graphing a division of two functions?

anomalousdecay · Mar 16, 2014

braintic said:
*************************************************

Regarding l'Hopital's rule .... at a recent conference on the marking of the 2013 HSC maths papers, they made it clear that any results that are beyond the NSW HSC syllabus (such as l'Hopital's rule) must be FULLY explained in order to use them. If you use l'Hopital's rule without any explanation you will get zero marks.

Regarding this, they should just make it part of the syllabus soon.

Sy123 · Mar 16, 2014

Trebla said:
If you're just sketching don't you just apply the principles of graphing a division of two functions?

The issue is the limit of the function at zero, which cannot be seen graphically

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