# Limit.

• Mar 1st 2009, 11:56 AM
Showcase_22
Limit.
Quote:

Find the value of $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}$.
This is what I did:

$\displaystyle f(x)=sin(x)-x$
$\displaystyle f'(x)=cos(x)-1$
$\displaystyle f^{(2)}(x)=-sin(x)$
$\displaystyle f^{(3)}(x)=-cos(x)$

$\displaystyle g(x)=tan(x)-x$
$\displaystyle g'(x)=sec^2(x)-1$
$\displaystyle g^{(2)}(x)=\frac{2sin(x)}{cos^3(x)}$
$\displaystyle g^{(3)}(x)=\frac{2cos^2(x)+6sin^2(x)}{cos^4(x)}$

Hence:

$\displaystyle a_3=f^{(3)}(0)=-cos(0)=-1$
$\displaystyle b_3=g^{(3)}(0)=2$

Hence $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}=\frac{a_3}{b_3}=-\frac{1}{2}$

However, when I put x=0.0000000001 into my calculator (as a check) I end up with 1.

What's going wrong??? :S
• Mar 1st 2009, 12:06 PM
redsoxfan325
According to Maple 12, $\displaystyle -\frac{1}{2}$ is correct. Also, just looking at the graph of the function, it appears that $\displaystyle f(x) = \frac{sin(x)-x}{tan(x)-x}$ is continuous at $\displaystyle x = 0$. I think your mistake occurred when you plugged in 0.0000000001.
• Mar 1st 2009, 12:08 PM
Showcase_22
Oh right.

It's a rather strange function since putting 0.01 into the function seems to suggest it tends to 0, but putting something much smaller in seems to suggest it tends to 1!

Well I guess there's no substitute for analytical methods!

thanks!