Limit.

**Showcase_22** · March 1st 2009, 11:56 AM

Find the value of $\lim_{x \rightarrow 0} \frac{sin(x)-x}{tan(x)-x}$ .

This is what I did:

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

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

Hence:

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

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

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

What's going wrong??? :S

**redsoxfan325** · March 1st 2009, 12:06 PM

According to Maple 12, $-\frac{1}{2}$ is correct. Also, just looking at the graph of the function, it appears that $f(x) = \frac{sin(x)-x}{tan(x)-x}$ is continuous at $x = 0$ . I think your mistake occurred when you plugged in 0.0000000001.

**Showcase_22** · March 1st 2009, 12:08 PM

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!

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