1. ## 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?

2. 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

3. 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 ..so the book is wrong then for sure?

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

5. 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

6. 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.)

7. 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

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

8. 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.

9. 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

10. Hi
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 :

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