# Somebody check if this is ok!

• Jun 16th 2008, 05:57 AM
lynch-mob
Somebody check if this is ok!
$\displaystyle \lim_{x\rightarrow0}\frac{-42(ln(1+x))+42x}{x^2(e^x+4)}=$

$\displaystyle \frac{-42[x-\frac{x^2}{2}+\frac{x^3}{3}...]+42x}{x^2[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+4]}=$

$\displaystyle \frac{[21x^2-14x^3...]}{[5x^2+x^3+\frac{x^4}{2!}...]}$

Division with $\displaystyle x^2$ gives

$\displaystyle \frac{21-14x...}{5+x+\frac{x^2}{2!}...}=$ and when$\displaystyle x\rightarrow0$ we get

$\displaystyle \frac{21-0}{5+0+0}=\frac{21}{5}$??
• Jun 16th 2008, 06:37 AM
flyingsquirrel
Hi

That's it. :) By the way notice that the terms $\displaystyle x^3,\,x^4,\ldots$ are useless here :

$\displaystyle \frac{-42\ln(1+x)-42x}{x^2(\exp x+4)}=\frac{-42\left(x-\frac{x^2}{2}+o(x^2)\right)+42x}{x^2(1+o(1)+4)}=\f rac{21x^2+o(x^2)}{5x^2+o(x^2)} \to \frac{21}{5}$ as $\displaystyle x\to 0$
• Jun 16th 2008, 06:38 AM
TKHunny
1) You're missing some ellipses in that horrible 2nd denominator.

2) You physically substituted x = 0 in the last step, even though you divided by x^2 a few steps earlier. This should at least make you feel a little dirty.

3) I'm not particularly fond of the 3rd numerator. All the shown coefficients are integers. Very deceptive.

4) You seem to have the idea. Good work.

5) Can you think of another way to do it? This is useful for checking things for yourself.
• Jun 16th 2008, 06:41 AM
lynch-mob
thanx
Quote:

Originally Posted by flyingsquirrel
Hi

That's it. :) By the way notice that the terms $\displaystyle x^3,\,x^4,\ldots$ are useless here :

Thanks for your check. When do they become useless. After or before the division with $\displaystyle x^2$
• Jun 16th 2008, 07:46 AM
flyingsquirrel
Quote:

Originally Posted by lynch-mob
Thanks for your check. When do they become useless. After or before the division with $\displaystyle x^2$

Before the division : When you divide by $\displaystyle x^2$ you should see that these "useless" terms tend to 0 so they do not give us any information on the limit, they shouldn't appear at all. To find which terms have to be written, one can simply try several approximations starting from the one which has the lowest order :

Using $\displaystyle \ln x =x+o(x)$ gives $\displaystyle -42\ln (1+x)+42x=o(x)$ which doesn't give us any information on the limit since it is divided by $\displaystyle x^2(\exp x+4)$

Using $\displaystyle \ln x =x-\frac{x^2}{2}+o(x^2)$ gives $\displaystyle -42\ln (1+x)+42x=-\frac{x^2}{2}+o(x^2)$ which is enough to get the limit since one can give an approximation of the denominator with order two.
• Jun 16th 2008, 09:02 AM
Mathstud28
Quote:

Originally Posted by lynch-mob
$\displaystyle \lim_{x\rightarrow0}\frac{-42(ln(1+x))+42x}{x^2(e^x+4)}=$

$\displaystyle \frac{-42[x-\frac{x^2}{2}+\frac{x^3}{3}...]+42x}{x^2[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+4]}=$

$\displaystyle \frac{[21x^2-14x^3...]}{[5x^2+x^3+\frac{x^4}{2!}...]}$

Division with $\displaystyle x^2$ gives

$\displaystyle \frac{21-14x...}{5+x+\frac{x^2}{2!}...}=$ and when$\displaystyle x\rightarrow0$ we get

$\displaystyle \frac{21-0}{5+0+0}=\frac{21}{5}$??

A slight trick you could have done would be

$\displaystyle e^x+4=e^x-1+5\sim{x+5}$
It could have saved some writing