# Textbook discrepancy

• May 14th 2008, 07:31 PM
Mathstud28
Textbook discrepancy
Today in one of my texts I saw this

$\displaystyle \text{As}x\to{0},\tan(x^3)\sim{x^2}$

I dont think this is right?

Correct me if I am wrong but Three ways prove I am right unless I am missing somethign stupid

$\displaystyle \lim_{x\to{0}}\frac{\tan(x^3)}{x^2}=\lim_{x\to{0}} \frac{\frac{\sin(x^3)}{\cos(x^3)}}{x^2}=\lim_{x\to {0}}\frac{sin(x^3)}{\cos(x^3)\cdot{x^2}}$

Now I will apply that $\displaystyle \sin(x^3)\sim{x^3}\text{As}x\to{0}$

To rewrite this as $\displaystyle \lim_{x\to{0}}\frac{x^3}{\cos(x^3)\cdot{x^2}}=\lim _{x\to{0}}\frac{x}{\cos(x^3)}=0\ne{1}$

You can also do this limit through power series and L'hopital's.

Where am I making my mistake?
• May 14th 2008, 07:56 PM
topsquark
Quote:

Originally Posted by Mathstud28
Today in one of my texts I saw this

$\displaystyle \text{As}x\to{0},\tan(x^3)\sim{x^2}$

I dont think this is right?

Correct me if I am wrong but Three ways prove I am right unless I am missing somethign stupid

$\displaystyle \lim_{x\to{0}}\frac{\tan(x^3)}{x^2}=\lim_{x\to{0}} \frac{\frac{\sin(x^3)}{\cos(x^3)}}{x^2}=\lim_{x\to {0}}\frac{sin(x^3)}{\cos(x^3)\cdot{x^2}}$

Now I will apply that $\displaystyle \sin(x^3)\sim{x^3}\text{As}x\to{0}$

To rewrite this as $\displaystyle \lim_{x\to{0}}\frac{x^3}{\cos(x^3)\cdot{x^2}}=\lim _{x\to{0}}\frac{x}{\cos(x^3)}=0\ne{1}$

You can also do this limit through power series and L'hopital's.

Where am I making my mistake?

$\displaystyle tan(x^3) \approx x^3$ for small x. Just find the Maclaurin series. No heavy lifting with the limits is required.

-Dan
• May 14th 2008, 07:59 PM
Mathstud28
Quote:

Originally Posted by topsquark
$\displaystyle tan(x^3) \approx x^3$ for small x. Just find the Maclaurin series. No heavy lifting with the limits is required.

-Dan

Yeah that is what I said...and of course I did use Maclaurin (Rofl)..so the book is wrong then for sure?
• May 14th 2008, 08:07 PM
hatsoff
Quote:

Originally Posted by Mathstud28
$\displaystyle \sin(x^3)\sim{x^3}\text{As}x\to{0}$

• May 14th 2008, 08:11 PM
Mathstud28
Quote:

Originally Posted by hatsoff

This notation means that as $\displaystyle x\to{0}$ then $\displaystyle \sin(x^{n})\to{x^n}$

Really what it means is as x approaches 0 $\displaystyle \sin(x^n)$ and $\displaystyle x^{n}$ become equivalent
• May 14th 2008, 08:16 PM
hatsoff
Quote:

Originally Posted by Mathstud28
This notation means that as $\displaystyle x\to{0}$ then $\displaystyle \sin(x^{n})\to{x^n}$

Really what it means is as x approaches 0 $\displaystyle \sin(x^n)$ and $\displaystyle x^{n}$ become equivalent

Okay, so, in other words:

$\displaystyle \lim_{x\to{0}}\frac{sinx^3}{x^3}=1$

But how do you determine that it really does equal 1? (I don't know how to work with trig limits.)
• May 14th 2008, 08:20 PM
Mathstud28
Quote:

Originally Posted by hatsoff
Okay, so, in other words:

$\displaystyle \lim_{x\to{0}}\frac{sinx^3}{x^3}=1$

But how do you determine that it really does equal 1? (I don't know how to work with trig limits.)

Two ways

L'hopital's

$\displaystyle \lim_{x\to{0}}\frac{\sin(x^3)}{x^3}=\lim_{x\to{0}} \frac{\cos(x^3)\cdot{3x^2}}{3x^2}=\lim_{x\to{0}}\c os(x^3)=1$

Or inserting the Macluarin series you get

$\displaystyle \lim_{x\to{0}}\frac{x^3-\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}-...}{x^3}=\lim_{x\to{0}}1-\frac{x^{6}}{3!}+\frac{x^{12}}{5!}-...=1$
• May 14th 2008, 08:23 PM
hatsoff
Quote:

Originally Posted by Mathstud28
Two ways

L'hopital's

$\displaystyle \lim_{x\to{0}}\frac{\sin(x^3)}{x^3}=\lim_{x\to{0}} \frac{\cos(x^3)\cdot{3x^2}}{3x^2}=\lim_{x\to{0}}\c os(x^3)=1$

Or inserting the Macluarin series you get

$\displaystyle \lim_{x\to{0}}\frac{x^3-\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}-...}{x^3}=\lim_{x\to{0}}1-\frac{x^{6}}{3!}+\frac{x^{12}}{5!}-...=1$

Hmph. So that's twice in one night that Maclaurin series have reared their heads. I will have to make myself more familiar with them; presently I find them to be quite unwieldy, for lack of practice/use.

Thanks.
• May 14th 2008, 08:26 PM
Mathstud28
Quote:

Originally Posted by hatsoff
Hmph. So that's twice in one night that Maclaurin series have reared their heads. I will have to make myself more familiar with them; presently I find them to be quite unwieldy, for lack of practice/use.

Thanks.

I love Maclaurin series...I actually just created a long and hopefully enlightening tutorial on them today. I posted it like an hour ago but it might be gone for review
• May 14th 2008, 09:47 PM
flyingsquirrel
Hi
Quote:

Originally Posted by hatsoff
Hmph. So that's twice in one night that Maclaurin series have reared their heads. I will have to make myself more familiar with them; presently I find them to be quite unwieldy, for lack of practice/use.

If you don't know MacLaurin series, you can use the definition of the derivative of the sine taken at 0 :

$\displaystyle \lim_{x\to0}\frac{\sin x^3}{x^3}=\lim_{X\to0} \frac{\sin X-\sin 0}{X-0} =\sin'(0)=\cos 0=1$