1. ## trigonometric limit problem

lim (sin(x) - x) / x^3
x~>0

so i tried splitting it up into sin(x) / x^3 - x / x^3. sin(x) / x^3 = (1/x^2)(sin(x) / x) and the limit as x approaches 0 of sin(x)/x = 1 so the first term just becomes 1/x^2. for the second term i canceled out an x from the top and bottom so i also got 1/x^2. so 1/x^2 - 1/x^2 = 0, so i got that the limit is equal to 0. but when i used L'Hospital's Rule on it, i got a completely different answer, -1/6. was the algebraic way i was doing it wrong? please help.

2. Originally Posted by oblixps
lim (sin(x) - x) / x^3
x~>0

so i tried splitting it up into sin(x) / x^3 - x / x^3. sin(x) / x^3 = (1/x^2)(sin(x) / x) and the limit as x approaches 0 of sin(x)/x = 1 so the first term just becomes 1/x^2. for the second term i canceled out an x from the top and bottom so i also got 1/x^2. so 1/x^2 - 1/x^2 = 0, so i got that the limit is equal to 0. but when i used L'Hospital's Rule on it, i got a completely different answer, -1/6. was the algebraic way i was doing it wrong? please help.
You're disregarding an important limit property!

$\displaystyle \lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\cdot\lim_{x\to a} g(x)$

So, $\displaystyle \lim_{x\to0}\frac{\sin x-x}{x^3}=\lim_{x\to0}\frac{\sin x}{x}\cdot\frac{1}{x^2}-\lim_{x\to0}\frac{1}{x^2}$ $\displaystyle =\left[\lim_{x\to0}\frac{\sin x}{x}\lim_{x\to0}\frac{1}{x^2}\right]-\lim_{x\to0}\frac{1}{x^2}=(1)(\infty)-(\infty)=\infty-\infty$, an indeterminate case.

So applying L'Hôpital's rule would be advisible here -- the answer would be $\displaystyle -\tfrac{1}{6}$.

3. Hello, oblixps!

Your algebra is off . . .

$\displaystyle \lim_{x\to0}\frac{\sin x - x}{x^2}$

Presently, we have: .$\displaystyle \lim_{x\to0}\frac{\sin x - x}{x^3} \;\Rightarrow\; \frac{0}{0}$ . . . an indeterminate form.

We can write this as: .$\displaystyle \lim_{x\to0}\frac{\frac{\sin x}{x} - 1}{x^2} \;\Rightarrow\; \frac{1-1}{0^2} \:=\:\frac{0}{0}$ . . . an indeterminate form -- again!

We can apply L'Hopital: .$\displaystyle \lim_{x\to0}\frac{\cos x - 1}{3x^2} \;=\;\frac{1-1}{0} \:=\:\frac{0}{0}$ . . . indeterminate

At this point, we can: .$\displaystyle \begin{array}{cc}(1) & \text{Apply L'Hopital twice} \\ (2) & \text{Do some clever algebra} \end{array}$

Method 1:

. . L'Hopital again: .$\displaystyle \lim_{x\to0}\frac{-\sin x}{6x} \:=\:\frac{0}{0}$ . . . indeterminate

. . L'Hopital again: .$\displaystyle \lim_{x\to0}\frac{-\cos x}{6} \;=\;\boxed{-\frac{1}{6}}$

Method 2:

. . We have: .$\displaystyle \frac{\cos x - 1}{3x^2} \;=\;\frac{-(1-\cos x)}{3x^2}$

. . Multiply by $\displaystyle \frac{1+\cos x}{1+\cos x}\!:\quad \frac{-(1-\cos x)}{3x^2}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{-(1-\cos^2\!x)}{3x^2(1+\cos x)} \;=\;\frac{-\sin^2\!x}{3x^2(1+\cos x)}$ . $\displaystyle = \;\frac{-1}{3(1+\cos x)}\cdot\frac{\sin^2\! x}{x^2}$

. . Take the limit: .$\displaystyle \lim_{x\to0}\left[\frac{-1}{3(1+\cos x)}\cdot\left(\frac{\sin x}{x}\right)^2\right] \;=\;\frac{-1}{3(2)}\cdot(1)^2 \;=\;\boxed{-\frac{1}{6}}$

4. or expand $\displaystyle \sin x$ in a taylor series about $\displaystyle x=0$

then $\displaystyle \lim_{x \to 0} \frac{\sin x -x}{x^{3}} = \lim_{x \to 0} \frac{-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5} - \frac{1}{7!}x^{7}+ ...}{x^{3}}$

$\displaystyle = \lim_{x \to 0} \Big(-\frac{1}{3!}+\frac{1}{5!}x^{2} - \frac{1}{4!}x^{7}+ ... \Big) =$ $\displaystyle -\frac{1}{3!} = -\frac{1}{6}$

5. Originally Posted by Random Variable
or expand $\displaystyle \sin x$ in a taylor series about $\displaystyle x=0$

then $\displaystyle \lim_{x \to 0} \frac{\sin x -x}{x^{3}} = \lim_{x \to 0} \frac{-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5} - \frac{1}{7!}x^{7}+ ...}{x^{3}}$

$\displaystyle = \lim_{x \to 0} \Big(-\frac{1}{3!}+\frac{1}{5!}x^{2} - \frac{1}{4!}x^{7}+ ... \Big) =$ $\displaystyle -\frac{1}{3!} = -\frac{1}{6}$
I would've never thought of this approach. Thanks for the enlightenment.