A limit :s

1. Mar 2, 2006

mohlam12

hey again!
this time, I have this one to solve

$$\lim_{x\rightarrow 0}\frac{\tan(x)-\sin(x)}{x^3}$$

i went like this

$$\lim_{x\rightarrow 0}\frac{\frac{tan(x)}{x} - \frac{sin(x)}{x}}{x^2}$$

= lim (0/0)

which is always an undetermined form... is there any other way to solve this WITHOUT using derivatives (not learned yet)

Thank you!

Last edited: Mar 2, 2006
2. Mar 2, 2006

arildno

There's ALWAYS another way!
Here's how you could start out:
$$\lim_{x\to[0}}\frac{\tan(x)-\sin(x)}{x^{3}}=\lim_{x\to{0}}\frac{\sin(x)}{x}\frac{\frac{1}{\cos(x)}-1}{x^{2}}=\lim_{x\to{0}}\frac{\sin(x)}{x}\frac{1-\cos^{2}(x)}{\cos(x)(1+\cos(x))x^{2}}=\lim_{x\to{0}}(\frac{\sin(x)}{x})^{3}\frac{1}{\cos(x)(1+\cos(x))}$$
Can you take it from there?

3. Mar 2, 2006

mohlam12

I think so,
So the answer is 1/2 ?

4. Mar 2, 2006

arildno

You are not sure about that?

5. Mar 2, 2006

mohlam12

I am actually!
Thank you

6. Mar 2, 2006

arildno

You're welcome.