Textbook discrepancy

**Mathstud28** · May 14th 2008, 07:31 PM

Today in one of my texts I saw this

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

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

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

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

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

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

Where am I making my mistake?

**topsquark** · May 14th 2008, 07:56 PM

Originally Posted by Mathstud28

Today in one of my texts I saw this

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

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

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

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

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

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

Where am I making my mistake?

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

-Dan

**Mathstud28** · May 14th 2008, 07:59 PM

Originally Posted by topsquark

$tan(x^3) \approx x^3$ 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?

**hatsoff** · May 14th 2008, 08:07 PM

Originally Posted by Mathstud28

$\sin(x^3)\sim{x^3}\text{As}x\to{0}$

Please elaborate.

**Mathstud28** · May 14th 2008, 08:11 PM

Originally Posted by hatsoff

Please elaborate.

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

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

**hatsoff** · May 14th 2008, 08:16 PM

Originally Posted by Mathstud28

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

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

Okay, so, in other words:

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

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

**Mathstud28** · May 14th 2008, 08:20 PM

Originally Posted by hatsoff

Okay, so, in other words:

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

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

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

Or inserting the Macluarin series you get

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

**hatsoff** · May 14th 2008, 08:23 PM

Originally Posted by Mathstud28

Two ways

L'hopital's

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

Or inserting the Macluarin series you get

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

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.

**Mathstud28** · May 14th 2008, 08:26 PM

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

**flyingsquirrel** · May 14th 2008, 09:47 PM

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 :

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

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